123–129-Page Sears and Zemanski’s University Physics with Modern Physics Problems Solution Hugh D. Young & Roger A. Freedman | DSC Physics | Semester 1

 

Description:

This post provides the step-by-step solved problems from pages 123–129 of Sears and Zemanski’s University Physics with Modern Physics by Hugh D. Young & Roger A. Freedman. These solutions are specially prepared for DSC Physics, Semester 1 students, making it easier to understand complex concepts, master problem-solving techniques, and strengthen your foundation in physics.

The solutions cover a range of fundamental physics problems aligned with the textbook, ensuring that you not only get the correct answers but also a detailed explanation of the reasoning and methodology behind each step. Whether you are revising for exams, practicing numerical problems, or clarifying concepts you find challenging, these solutions will serve as an essential guide.

What you’ll find in this resource:

• Complete worked-out solutions from pages 123–129

• Clear explanations for each step in solving the problems

• Useful for DSC Physics (Semester 1) coursework and exam preparation

• Based on the authoritative textbook by Young & Freedman

• Helps in bridging the gap between theory and application in physics

This post is designed for university-level physics students who are following Sears & Zemanski’s University Physics with Modern Physics. It will not only help you in solving numerical problems but also improve your conceptual clarity, enabling you to tackle similar problems on your own.

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Here you go—each answer is paired with its question and number.

Q4.1 Can a body be in equilibrium when only one force acts on it? Explain.
Answer: No. Equilibrium requires zero net force. A single nonzero force cannot sum to zero, so the body must accelerate.

Q4.2 A ball thrown straight up has zero velocity at its highest point. Is the ball in equilibrium at this point? Why or why not?
Answer: No. Although the velocity is momentarily zero, gravity still provides a downward force, so the acceleration is not zero—hence not in equilibrium.

Q4.3 A helium balloon hovers in midair, neither ascending nor descending. Is it in equilibrium? What forces act on it?
Answer: Yes—static equilibrium. Upward buoyant force equals downward weight (and any drag is negligible when it’s at rest).

Q4.4 When you fly in an airplane at night in smooth air, you have no sensation of motion, even though the plane may be moving at 800 km/h. Why?
Answer: Your body senses acceleration, not constant velocity. With smooth (≈zero-acceleration) flight and no visual cues, you feel “at rest.”

Q4.5 If the two ends of a rope in equilibrium are pulled with forces of equal magnitude and opposite directions, why isn’t the total tension in the rope zero?
Answer: Tension is the magnitude of the internal pulling force transmitted along the rope, not a vector sum of the pulls at opposite ends. Each cross-section carries a nonzero tensile force even though the net external force on the rope can be zero.

Q4.6 You tie a brick to a rope and whirl it in a horizontal circle. Describe the path after you suddenly let go.
Answer: It moves in a straight line tangent to the circle at the release point, then follows a projectile (parabolic) path downward under gravity.

Q4.7 When a car stops suddenly, passengers tend to move forward; when it makes a sharp turn, they slide to one side. Why?
Answer: Inertia. Your body keeps its state of motion. When the seat/car decelerates or turns (accelerates sideways), your body lags, so you move forward or sideways relative to the car until forces from the seat/seatbelt change your motion.

Q4.8 Some say a “force of inertia” throws passengers forward when a car brakes. What’s wrong with this?
Answer: Inertia isn’t a force; it’s the tendency to maintain velocity. In an accelerating (non-inertial) car frame you may introduce a fictitious “inertial force” for bookkeeping, but physically the forward motion occurs because no real backward force acts until the seatbelt/seat provides one.

Q4.9 In a windowless moving bus, a ball at rest in the aisle starts to move toward the rear. Give two explanations and a way to decide which is correct.
Answer: (1) The bus accelerated forward, making loose objects appear to slide toward the rear in the bus frame. (2) The bus is on a downward-tilted floor (road pitched so gravity has a component toward the rear). Test: hang a small pendulum—if it tilts backward while the bus feels level, that indicates forward acceleration; if it hangs perpendicular to the floor yet the ball rolls, the floor is tilted.

Q4.10 If force, length, and time are fundamental quantities, what are the units of mass?
Answer: From $F=ma$, $m = F\,T^{2}/L$. So the unit of mass is (unit of force)·s²/m (e.g., N·s²·m⁻¹).

Q4.11 Why is the Earth only approximately an inertial reference frame?
Answer: Earth rotates and orbits, producing small accelerations (e.g., centrifugal and Coriolis effects ~0.3–0.4% of $g$ at the equator) and its gravity field is nonuniform. For many lab situations these are tiny, so Earth is an approximate inertial frame.

Q4.12 Does Newton’s second law hold for an observer in a van as it speeds up, slows down, or rounds a corner? Explain.
Answer: Not in its simple form. Those are accelerating (non-inertial) motions; to use $\sum \mathbf{F} = m\mathbf{a}$ in the van’s frame you must add fictitious forces. In a van moving at constant velocity on a straight road, no fictitious forces are needed.

Q4.13 Some students call $m\mathbf{a}$ “the force of acceleration.” Is that correct?
Answer: No. $m\mathbf{a}$ equals the net real force acting on the mass (by Newton’s second law); it isn’t an additional force. Better: call it the “required net force” to produce acceleration $\mathbf{a}$. (In non-inertial analysis, one sometimes introduces a fictitious inertial force $-m\mathbf{a}_{\text{frame}}$, but that’s a modeling convenience.)

Q4.14 The acceleration of a falling body is measured in an elevator traveling upward at a constant speed of 9.8 m/s. What value is obtained?
Answer: $9.8\ \text{m/s}^2$ downward (≈$g$). Constant speed means no elevator acceleration, so the frame is effectively inertial.

Q4.15 Can you play normal catch in a bus moving at constant speed while turning on a level road? Why or why not?
Answer: Not “just like at rest.” During a turn there is centripetal acceleration; thrown balls travel straight in the ground frame but the bus curves underneath them, so paths appear deflected. You must aim into the turn to compensate.

Q4.16 Students sometimes say the force of gravity on an object is $9.8\ \text{m/s}^2$. What’s wrong?
Answer: $9.8\ \text{m/s}^2$ is acceleration, not force. The gravitational force (weight) is $W = mg$ and has units of newtons.

Q4.17 Why can it hurt more to kick a big rock than a small pebble? Must the big rock hurt more? Explain.
Answer: A massive rock accelerates very little, so your foot’s velocity changes quickly (short stopping time), producing a large contact force on your foot (Newton’s 3rd & impulse). A small pebble accelerates away, reducing the force. It need not always hurt more—shoes, kicking gently, or a compliant surface can lengthen stopping time and reduce force; a sharp pebble could cause more localized injury.

Q4.18 “It’s not the fall that hurts you; it’s the sudden stop at the bottom.” Translate using Newton’s laws.
Answer: Injury arises from the large deceleration over a very short stopping time, which creates a large average force (impulse–momentum, $J=\Delta p = \bar{F}\,\Delta t$; work–energy via short stopping distance → large force).

Q4.19 Why can a person dive from 10 m into water without injury but be seriously hurt landing on concrete from the same height?
Answer: Water allows your body to decelerate over a longer distance/time (you penetrate and can streamline), reducing force. Concrete stops you almost instantly—tiny stopping distance/time → very large force and pressure on tissues.

Nice — here are Q4.20 → Q4.32 with each question followed by its answer and explanation.

Q4.20 Why are cars designed to crumple in front and back for safety? Why not for side collisions and rollovers?
Answer: Crumple zones increase the distance and time over which the car’s kinetic energy and the passengers’ momentum are brought to zero, reducing peak forces on occupants (work–energy / impulse). Front and rear crumple zones have space to deform and absorb energy. Side collisions and rollovers offer much less structural “crush” space beside/above occupants (the cabin is narrower/closer), so protection focuses on stiff passenger cell, reinforced doors, side airbags, and strong roof structure—design tradeoffs between energy absorption and preventing intrusion.

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Q4.21 When a string barely strong enough lifts a heavy weight can lift it by a steady pull; but if you jerk the string, it will break. Explain in terms of Newton’s laws.
Answer: A jerk produces a much larger instantaneous acceleration (large $\Delta v/\Delta t$), so the tension briefly spikes well above the steady value ( $T = m(a+g)$ in vertical lift ), possibly exceeding the string’s breaking strength. Steady pull gives moderate, sustained tension; a jerk creates a transient high tension.

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Q4.22 A large crate is suspended from the end of a vertical rope. Is the tension greater when the crate is at rest or when it is moving upward at constant speed? If the crate is traveling upward, is the tension greater when the crate is speeding up or when it is slowing down?
Answer:

• At rest vs moving upward at constant speed: tension is the same in both cases — it equals the crate’s weight (no net acceleration).

• When speeding up upward: tension is greater than the weight (net upward acceleration $a$, so $T = m(g + a)$).

• When slowing down while moving upward (i.e., acceleration downward): tension is less than the weight ($T = m(g - |a|)$).

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Q4.23 Which feels a greater pull due to the earth’s gravity: a 10-kg stone or a 20-kg stone? If you drop the two stones, why doesn’t the 20-kg stone fall with twice the acceleration of the 10-kg stone?
Answer: The 20-kg stone feels a greater gravitational force (twice the magnitude) because $F_g = mg$. But both fall with the same acceleration $g$ because $a = F_{\text{net}}/m = mg/m = g$ — the mass cancels out.

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Q4.24 Why is it incorrect to say that 1.0 kg equals 2.2 lb?
Answer: The statement conflates mass and force units. 1.0 kilogram (kg) is a mass; 2.20462 "pounds mass" (lbm) is the equivalent mass in imperial units — so saying “1.0 kg ≈ 2.2 lbm” is acceptable if you mean pound-mass. But a pound (lbf) often denotes force; 1 kg under Earth gravity weighs about 9.8 N, which equals about 2.2046 lbf. So be explicit: 1.0 kg ≈ 2.20462 lbm (mass). 1.0 kg corresponds to ≈2.20462 lbf only if you treat lbf as the weight of 1 lbm on Earth.

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Q4.25 A horse is hitched to a wagon. Since the wagon pulls back on the horse just as hard as the horse pulls on the wagon, why doesn’t the wagon remain in equilibrium, no matter how hard the horse pulls?
Answer: Action–reaction pairs act on different bodies, so they don’t cancel on one object. The horse pulls the wagon forward; the wagon pulls the horse backward. For the system (horse+wagon) the internal pair cancels, but the horse also pushes the ground backward with its hooves and the ground pushes the horse forward (frictional reaction). That ground reaction provides the unbalanced forward force that moves the system.

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Q4.26 True or false? You exert a push $P$ on an object and it pushes back on you with a force $F$. If the object is moving at constant velocity, then $F$ is equal to $P$, but if the object is being accelerated, then $P$ must be greater than $F$.
Answer: True. By Newton’s third law $F$ and $P$ are equal in magnitude (they are the action–reaction pair). For the object to accelerate, the net force on it must be nonzero: $P - (\text{other opposing forces}) = ma$. So when accelerating, the external opposing forces (e.g., friction, drag, whatever) differ, and the push $P$ must exceed those opposing forces; the reaction force on you from the object remains equal in magnitude to $P$.

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Q4.27 A large truck and a small compact car have a head-on collision. During the collision, the truck exerts a force $F_{T\text{ on }C}$ on the car, and the car exerts a force $F_{C\text{ on }T}$ on the truck. Which force has the larger magnitude, or are they the same? Does your answer depend on how fast each vehicle was moving? Why or why not?
Answer: The magnitudes are the same at every instant (Newton’s third law). This does not depend on the pre-collision speeds. What does differ are the resulting accelerations: the smaller mass experiences a larger acceleration for the same force ($a = F/m$).

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Q4.28 When a car comes to a stop on a level highway, what force causes it to slow down? When the car increases its speed on the same highway, what force causes it to speed up?
Answer: To stop: braking causes the wheels to exert a backward force on the road (through tire friction) and the road exerts an equal-and-opposite forward force on the car? Wait — clearer: Braking makes the wheels try to rotate so the contact tends to oppose forward motion; frictional (static/kinetic) forces between tires and road act backward on the car (plus air resistance), producing deceleration. To speed up: the engine turns the wheels pushing the road backward; static friction from the road acts forward on the tires, producing the forward force that accelerates the car.

(Short version: the friction between tires and road provides the backward force to decelerate under braking and the forward force to accelerate when the engine drives the wheels.)

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Q4.29 A small compact car is pushing a large van that has broken down, and they travel with equal velocities and accelerations. While the car is speeding up, is the force it exerts on the van larger than, smaller than, or the same magnitude as the force the van exerts on it? Which vehicle has the larger net force on it, or are the net forces the same?
Answer: The force the car exerts on the van equals in magnitude the force the van exerts on the car (third law). Net forces differ: net force on each = mass × common acceleration. Because the van has larger mass, its net force $m_{\rm van}a$ is larger than the car’s net force $m_{\rm car}a$. Those net forces arise from different balances of engine thrust, ground friction, and the interaction force between vehicles.

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Q4.30 Consider a tug-of-war between two people who pull in opposite directions on the ends of a rope. By Newton’s third law, the force that A exerts on B is just as great as the force that B exerts on A. So what determines who wins?
Answer: Winning depends on each person’s net horizontal force — which depends on how large a horizontal reaction they can get from the ground (static friction under their feet) and how they apply internal forces. Free-body diagrams show for each person: tension from the rope (backward or forward), friction from the ground, normal, weight. The person who can produce a larger static frictional force with the ground (better footing, more normal force, lower chance to slip) can produce the larger net forward force and thus win.

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Q4.31 Boxes A and B are in contact on a horizontal, frictionless surface. You push on box A with a horizontal 100-N force. Box A weighs 150 N, and box B weighs 50 N. Is the force that box A exerts on box B equal to 100 N, greater than 100 N, or less than 100 N? Explain.
Answer: Less than 100 N. Treat A+B as a system: total weight = 200 N → total mass = $200/g$. The acceleration is $a=F/(m_A+m_B)=100/((150+50)/g)=100g/200=0.5g\approx4.9\ \text{m/s}^2$. The contact force on B equals $m_B a$. With $m_B=50/g$, contact force $= (50/g)\cdot(0.5g)=25\ \text{N}$. So the push transmitted to B is 25 N.

Q4.32 A manual for student pilots: “When an airplane flies at a steady altitude, neither climbing nor descending, the upward lift force from the wings equals the plane’s weight. When the plane is climbing at a steady rate, the upward lift is greater than the weight; when the plane is descending at a steady rate, the upward lift is less than the weight.” Are these statements correct? Explain.
Answer: The statements are an oversimplification but are commonly used in elementary explanations. More precisely: for steady (constant-speed) flight there is no net acceleration, so the sum of vertical components of all aerodynamic and thrust forces equals the weight. In straight, level flight with thrust horizontal, lift ≈ weight. In a steady climb or descent at constant speed, the vertical components of lift and thrust together balance weight — so lift alone need not be greater (in a climb) or less (in a descent) than weight if thrust has a vertical component. Thus the manual’s statements are often approximately true for small climb/descent angles (and when thrust is nearly horizontal), but the correct condition is that vertical forces (lift + vertical thrust component) sum to weight in steady climb/descent.

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Nice — here are Q4.33 → Q4.38 and the three exercises. Each answer is numbered to match the question.

Q4.33 If your hands are wet and no towel is handy, you can remove some of the excess water by shaking them. Why does this work?
Answer: Shaking your hands gives the water droplets a rapid acceleration away from the skin. A droplet on your hand is held by adhesive/cohesive and by the contact force from the skin; when you shake, the droplet’s inertia resists the sudden change of motion of the hand and the droplet tends to continue moving in the (approximately) tangential direction. If the shake produces a large enough outward acceleration, the net force on the droplet (inertia vs. adhesion) becomes directed away from the hand and the droplet detaches and flies off. In short: shaking produces sufficient centripetal/centrifugal-like acceleration to overcome surface adhesion and gravity, so droplets separate.

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Q4.34 If you squat down and then suddenly get up, you may temporarily feel light-headed. What do Newton’s laws say about why this happens?
Answer: When you stand up suddenly your body (and the blood in it) is accelerated upward. The blood in your head has inertia and momentarily “lags” behind the motion of your body, so relative to your skull less blood is present in the brain for a short time → reduced cerebral blood (and oxygen) → light-headedness. Equivalently, your rapid upward acceleration produces a transient imbalance between forces (pressure + gravity) on the blood; until circulation readjusts you feel light.

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Q4.35 When a car is hit from behind, occupants may experience whiplash. Use Newton’s laws to explain.
Answer: In a rear impact the car is accelerated forward suddenly. The car’s seat pushes the torso forward quickly, but the head (which is not in direct contact with the seat) lags due to inertia. This relative motion causes the neck to experience a rapid backward–then–forward motion (hyperextension followed by flexion), producing strain/whiplash. Newton’s first law (inertia) explains the head’s tendency to remain in its original state while the car’s body moves.

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Q4.36 In a head-on collision, passengers not wearing seat belts may be thrown through the windshield. Why?
Answer: In a head-on collision the car decelerates very rapidly. Passengers’ bodies tend to keep moving forward (inertia). If nothing (or only soft contact) stops them, they continue forward at the pre-impact speed while the car stops around them — they therefore collide with the windshield or are ejected. Seat belts provide an external (restraining) force distributed over the torso to change the passenger’s momentum more safely.

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Q4.37 Head-on collision: compact 1000-kg car vs large 2500-kg car. Which experiences the greater force? Which greater acceleration? Why are small-car passengers more likely to be injured?
Answer:

• Force: During the collision, at any instant the magnitudes of the forces on the two cars are equal (Newton’s third law). Neither car experiences a larger contact force — the force pair is the same for both.

• Acceleration: The compact car experiences the larger acceleration because $a=F/m$. For the same force $F$, the smaller mass gives larger $a$ (i.e., $a_{\text{small}} = F/1000$ vs $a_{\text{large}} = F/2500$).

• Why more injury in small car: Greater acceleration (and usually greater deceleration over shorter stopping distances inside the passenger compartment) produces larger forces on occupants and more rapid changes of momentum, increasing injury risk. Also smaller cars typically have less crumple zone and less mass to absorb energy, so occupants endure larger decelerations.

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Q4.38 In a windowless rocket in deep space, how to tell (a) constant 0.8c motion vs (b) forward acceleration, without external contact?
Answer:
(a) If the rocket is moving at constant velocity (even 0.8c), it is an inertial frame — all local mechanical experiments behave as if at rest. You cannot detect uniform motion by internal experiments (principle of relativity). So you cannot tell whether you are coasting at 0.8c or at rest by internal experiments alone.
(b) If the rocket is accelerating forward, you can detect it internally: you will feel a continuous force pressing you “down” (analogous to gravity). Examples: an accelerometer will register a nonzero reading; loose objects will slide in a preferred direction; a plumb-line or suspended mass will hang tilted/deflected. In short: a nonzero reading on an accelerometer or persistent apparent gravity inside the cabin indicates acceleration. (This is true in both Newtonian and relativistic mechanics — acceleration is absolute and detectable locally.)

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Now the exercises.

E4.1 Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is $60.0^\circ$. If Rover exerts $F_1=270\ \text{N}$ and Fido exerts $F_2=300\ \text{N}$, find the magnitude of the resultant force and the angle it makes with Rover’s rope.
Solution (stepwise arithmetic shown):
Resultant magnitude for two vectors with angle $\theta$ between them:

$$R=\sqrt{F_1^2+F_2^2+2F_1F_2\cos\theta}.$$

Substitute $F_1=270$, $F_2=300$, $\theta=60^\circ$ ($\cos60^\circ=0.5$):

$$R=\sqrt{270^2+300^2+2(270)(300)(0.5)}.$$

Compute:

$$270^2=72{,}900,\quad 300^2=90{,}000,\quad 2(270)(300)(0.5)=81{,}000.$$

Sum: $72{,}900 + 90{,}000 + 81{,}000 = 243{,}900.$

$$R=\sqrt{243{,}900}\approx 493.86\ \text{N}.$$

Angle $\phi$ of resultant measured from Rover’s rope (the direction of $F_1$) can be found from components or the formula

$$\tan\phi=\dfrac{F_2\sin\theta}{F_1+F_2\cos\theta}.$$

With $\sin60^\circ=\sqrt{3}/2\approx0.8660254$, $\cos60^\circ=0.5$:

$$\tan\phi=\dfrac{300(0.8660254)}{270+300(0.5)}=\dfrac{259.8076}{270+150}=\dfrac{259.8076}{420}\approx0.61859,$$ $$\phi=\arctan(0.61859)\approx31.74^\circ.$$

Answer E4.1: Resultant magnitude $\approx 4.94\times10^{2}\ \text{N}$ (≈493.9 N), making about $31.7^\circ$ from Rover’s rope toward Fido’s rope.

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E4.2 Three horizontal ropes produce force vectors shown in Fig. E4.2. (a) Find the x and y components of each pull. (b) Use components to find the magnitude and direction of the resultant.
Solution / Method (general):
I don’t have the exact numeric values and orientations of the three forces from your figure (the text shows numbers 788 N and 411 N and angles 53° and 32°, but the third force’s magnitude/direction or sign isn’t unambiguously given here). Rather than guess the geometry, here is the clear method you should apply to the actual figure:

For each force $F_i$ with magnitude $F_i$ and angle $\theta_i$ measured from the positive $x$-axis (take care whether angle is above or below $x$-axis):

• $F_{i,x}=F_i\cos\theta_i$

• $F_{i,y}=F_i\sin\theta_i$ (take sign +/− appropriately)

Then sum components:

• $R_x=\sum_i F_{i,x}$

• $R_y=\sum_i F_{i,y}$

Resultant magnitude and direction:

$$R=\sqrt{R_x^2+R_y^2},\qquad \tan\Phi=R_y/R_x,$$

with $\Phi=$ angle of the resultant measured from +x (adjust quadrant with sign of $R_x,R_y$).

If you’d like, paste the three magnitudes and exactly how the angles are measured (above/below x or measured from y), and I’ll compute numeric components and the resultant for you.

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E4.3 Jaw strap: strap produces net upward force $5.00\ \text{N}$ on the chin. Tension same throughout the strap. To what tension must the strap be adjusted to provide the necessary upward force?
Solution / Method (common symmetric case):
The standard setup: strap passes under the chin and goes up on both sides making the same angle to the vertical. Let the tension in the strap be $T$. Each side provides an upward component; the two upward components add to the net upward force. If the angle between each side and the vertical is $\theta$ (i.e., the strap segments make angle $\theta$ from vertical), the upward component from one side is $T\cos\theta$. So

$$2T\cos\theta=5.00\ \text{N}\quad\Rightarrow\quad T=\dfrac{5.00}{2\cos\theta}.$$

If instead the angle given in your figure is measured from the horizontal (call that $\phi$), then the upward component of each side is $T\sin\phi$ and

$$2T\sin\phi=5.00\ \text{N}\quad\Rightarrow\quad T=\dfrac{5.00}{2\sin\phi}.$$

Answer E4.3: The required tension depends on the strap angle. For the common convention (angle $\theta$ from vertical): $T=5.00/(2\cos\theta)$. If you tell me the angle shown in the figure, I will compute the numeric tension.

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Below are clear, numbered answers for the problems you posted (I kept each item labeled with the same question number you used). Where a figure or graph is required but not provided I’ll note that and tell you what I need to finish the calculation.

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E4.4 A man drags a trunk up a ramp. Ramp slope = $20.0^\circ$. He pulls with a force $F$ whose direction makes $30.0^\circ$ with the ramp.
(a) How large must $F$ be so that the component parallel to the ramp (call it $F_x$) is $90.0\ \text{N}$?
(b) What is the perpendicular component $F_y$ then?

Solution & answers
(a) $F_x = F\cos(30^\circ)=90.0\ \text{N}$ so

$$F=\frac{90.0}{\cos30^\circ}=\frac{90.0}{0.8660254}\approx 103.9\ \text{N}.$$

(b) $F_y = F\sin(30^\circ)=103.9\times 0.5 = 51.94\ \text{N}$ (upward, perpendicular to ramp).

(Values rounded sensibly: $F\approx103.9\ \text{N},\;F_y\approx51.9\ \text{N}$.)

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4.22 The upward normal force exerted by the floor is $620\ \text{N}$ on an elevator passenger who weighs $650\ \text{N}$. What are the reaction forces to these two forces? Is the passenger accelerating? If so, find magnitude & direction of acceleration.

Answer & explanation

• Reaction to the normal force $N=620\ \text{N}$ (floor on passenger upward) is the force of the passenger on the floor, which is $620\ \text{N}$ downward.

• Reaction to the weight (Earth on passenger, $650\ \text{N}$ downward) is the force of the passenger on the Earth, $650\ \text{N}$ upward.

Net force on passenger: $F_{\text{net}}=N - W = 620 - 650 = -30\ \text{N}$ (negative = downward).
Passenger mass $m = W/g = 650/9.8 \approx 66.33\ \text{kg}$.
Acceleration $a = F_{\text{net}}/m = -30/66.33 \approx -0.452\ \text{m/s}^2$.

So the passenger is accelerating downward at about 0.452 m/s².

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4.5 Two forces act at a point: $F_1=9.00\ \text{N}$, direction $60.0^\circ$ above the x-axis in the second quadrant; $F_2=6.00\ \text{N}$, direction $53.1^\circ$ below the x-axis in the third quadrant. (a) Find x- and y-components of the resultant. (b) Magnitude of resultant.

Assumption / interpretation: “$60^\circ$ above x-axis in the second quadrant” means $F_1$ points at $180^\circ-60^\circ=120^\circ$ measured from +x. “$53.1^\circ$ below x-axis in the third quadrant” means $F_2$ points at $180^\circ+53.1^\circ=233.1^\circ$.

Components (to 3 sig figs):
Compute components:

• $F_{1x}=9.00\cos120^\circ=-4.50\ \text{N}$

• $F_{1y}=9.00\sin120^\circ=+7.79\ \text{N}$

• $F_{2x}=6.00\cos233.1^\circ\approx -3.59\ \text{N}$

• $F_{2y}=6.00\sin233.1^\circ\approx -4.80\ \text{N}$

Resultant components:

$$R_x = F_{1x}+F_{2x}\approx -4.50 + (-3.59) = -8.09\ \text{N},$$ $$R_y = F_{1y}+F_{2y}\approx 7.79 + (-4.80) = +2.99\ \text{N}.$$

(b) Magnitude:

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-8.09)^2+(2.99)^2}\approx 8.58\ \text{N}.$$

(Direction: $\tan\theta = R_y/R_x \Rightarrow\theta$ is in quadrant II; $\theta\approx \arctan(2.99/ -8.09)\approx 159.6^\circ$ from +x.)

(Values rounded: $R_x\approx -8.09\ \text{N},\;R_y\approx +2.99\ \text{N},\;R\approx8.58\ \text{N}.$)

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4.6 Electron ($m=9.11\times10^{-31}$ kg) starts at rest, travels $1.80\ \text{cm}=0.0180\ \text{m}$ and reaches speed $v=3.00\times10^{6}\ \text{m/s}$ under constant acceleration. Compute (a) acceleration, (b) time to reach grid, (c) net force.

Solution & answers
(a) $a = \dfrac{v^2- v_0^2}{2x} = \dfrac{(3.00\times10^6)^2}{2(0.0180)}$.
Numerically $a \approx \dfrac{9.00\times10^{12}}{0.036} = 2.50\times10^{14}\ \text{m/s}^2.$

(b) $t = v/a = (3.00\times10^6)/(2.50\times10^{14}) \approx 1.20\times10^{-8}\ \text{s}$ (12 ns).

(c) $F = m a = 9.11\times10^{-31}\times 2.50\times10^{14} \approx 2.28\times10^{-16}\ \text{N}.$

(Answers: $a\approx2.50\times10^{14}\ \text{m/s}^2,\; t\approx1.20\times10^{-8}\ \text{s},\; F\approx2.28\times10^{-16}\ \text{N}.$)

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4.7 A 68.5-kg skater initially moving at $2.40\ \text{m/s}$ comes to rest uniformly in $3.52\ \text{s}$ due to friction. What force does friction exert?

Solution & answer
Acceleration $a=(0-2.40)/3.52\approx -0.6818\ \text{m/s}^2$.
Friction force $F_{\text{fric}}=m a=68.5\times(-0.6818)\approx -46.7\ \text{N}$.

(The frictional force is about 46.7 N opposite to motion.)

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4.8 You step on a scale in an elevator. Your normal weight at rest is $625\ \text{N}$. (a) Elevator accelerates upward at $2.50\ \text{m/s}^2$. What does scale read? (b) If you hold a $3.85\ \text{kg}$ package by a light vertical string, what is the tension in the string during this acceleration?

Solution & answers
First find your mass: $m = 625/9.8 \approx 63.78\ \text{kg}.$

(a) Scale reads the normal force $N = m(g+a) = 63.78(9.8+2.5)=63.78(12.3)\approx 784.4\ \text{N}.$
So the scale reads ≈784 N.

(b) Tension in package string $T = m_{\text{pkg}}(g+a) = 3.85(9.8+2.5)=3.85(12.3)\approx 47.36\ \text{N}.$
So $T\approx 47.4\ \text{N}$ upward.

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4.9 Box on frictionless pond: applied horizontal force $48.0\ \text{N}$ causes $2.20\ \text{m/s}^2$. What is mass?

Answer
$m = F/a = 48.0/2.20 = 21.818\ \text{kg}$.
So $m\approx 21.82\ \text{kg}.$

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4.10 Worker pushes block with $80.0\ \text{N}$. Block starts from rest and moves $11.0\ \text{m}$ in $5.00\ \text{s}$. Friction negligible. (a) mass? (b) After worker stops at $t=5\ \text{s}$, how far in next $5\ \text{s}$?

Solution & answers
(a) From $s=\tfrac12 a t^2$, $a = 2s/t^2 = 2(11.0)/25 = 0.88\ \text{m/s}^2$. Then $m = F/a = 80.0/0.88 \approx 90.91\ \text{kg}.$

(b) Velocity at $t=5$: $v = a t = 0.88\times5 = 4.40\ \text{m/s}$. If pushing stops and friction ≈ 0, block moves with constant $v$ for the next 5 s, so displacement $= v\times5 = 4.40\times5 = 22.0\ \text{m}.$

(Answers: $m\approx90.9\ \text{kg},$ next-5-s distance $22.0\ \text{m}$.)

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4.11 Hockey puck $m=0.160\ \text{kg}$ at rest at $x=0$. From $t=0$ to $t=2.00\ \text{s}$ a constant force $0.250\ \text{N}$ acts. (a) Position and speed at $t=2.00\ \text{s}$. (b) If same force applied again from $t=5.00$ to $t=7.00\ \text{s}$, what are position and speed at $t=7.00\ \text{s}$?

Solution & answers
Acceleration $a=F/m = 0.250/0.160 = 1.5625\ \text{m/s}^2.$

(a) At $t=2.00$: $x=\tfrac12 a t^2 = 0.5(1.5625)(4)=3.125\ \text{m}$. Speed $v = a t = 1.5625\times2 = 3.125\ \text{m/s}.$

(b) From $t=2$ to $t=5$ no force → moves at constant $3.125\ \text{m/s}$ for 3 s → adds $9.375\ \text{m}$ so $x(5)=3.125+9.375=12.500\ \text{m}.$
Apply force again for 2 s (5→7): displacement during this interval = $v_0 t + \tfrac12 a t^2 = 3.125(2)+0.5(1.5625)(4)=6.25+3.125=9.375\ \text{m}.$
So $x(7)=12.500+9.375=21.875\ \text{m}.$ Speed at 7 s: $v(7)=v_0 + a t = 3.125 + 1.5625(2)=6.25\ \text{m/s}.$

(Answers rounded: $x(2)=3.125\ \text{m}, v(2)=3.125\ \text{m/s};\; x(7)=21.875\ \text{m}, v(7)=6.25\ \text{m/s}.$)

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4.12 Crate mass $32.5\ \text{kg}$ initially at rest is acted on by net horizontal force $14.0\ \text{N}$. (a) acceleration? (b) distance in $10.0\ \text{s}$? (c) speed at end of $10.0\ \text{s}$?

Solution & answers
(a) $a = F/m = 14.0/32.5 \approx 0.43077\ \text{m/s}^2.$
(b) $s = \tfrac12 a t^2 = 0.5(0.43077)(100) \approx 21.54\ \text{m}.$
(c) $v = a t = 0.43077\times10 \approx 4.3077\ \text{m/s}.$

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4.13 A 4.50-kg cart undergoes an acceleration in a straight line; the graph in Fig. E4.13 shows $a(t)$. (a) Find the maximum net force on this cart and when it occurs. (b) During what times ...?

Answer / note: I can’t compute numeric answers because the acceleration-vs-time graph (Fig. E4.13) was not supplied. To finish (a) and (b) I need the graph or a table of $a(t)$ values. If you paste or describe the acceleration segments (times and acceleration values), I will compute (i) $F_{\max}=m a_{\max}$ and (ii) the requested time-interval answers immediately.

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Here are clear, numbered answers for Q4.14 → Q4.21. Where a figure/graph is required but not provided I state that and give the exact steps you (or I) would use to get the numeric answer.

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Q4.14 A 2.75-kg cat moves in a straight line (x-axis). Figure E4.14 shows the x-component of velocity vs time.
(a) Find the maximum net force on the cat and when it occurs.
(b) When is the net force zero?
(c) What is the net force at $t=8.5\ \text{s}$?

Answer / method (graph required):
I don’t have the velocity-vs-time graph (Fig. E4.14) in your message, so I cannot give the numeric forces. Here’s exactly how to get the answers from the graph:

1. From the $v_x(t)$ graph compute the acceleration as the slope: $a_x(t)=dv_x/dt$. For piecewise linear graphs, find the slope of each line segment. (Units: m/s per s → m/s².)

2. Multiply by the mass to get net force: $F_x(t)=m\,a_x(t)$ with $m=2.75\ \text{kg}$.

3. (a) The maximum net force is $m$ times the maximum acceleration (largest positive slope) — report the value and the time where that slope occurs.

4. (b) Net force is zero wherever the slope $dv_x/dt$ equals zero (velocity is constant there).

5. (c) Read the slope at $t=8.5\ \text{s}$, multiply by 2.75 kg to get $F_x(8.5)$.

If you paste the graph data points or a small table of $v_x(t)$ values (or describe the segments with slopes and time intervals), I’ll compute all numeric answers immediately.

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Q4.15 An 8.00-kg rocket: $F(t)=A + B t^2$. Given $F(0)=100.0\ \text{N}$ and $F(2.00)=150.0\ \text{N}$. (a) Find $A,B$ with units. (b) Find net force and acceleration (i) at $t=0^+$ and (ii) at $t=3.00\ \text{s}$. (c) In deep space (no gravity), what is the acceleration at $t=3.00\ \text{s}$?

Answer (stepwise):
(a) At $t=0$: $F(0)=A+ B(0)^2 = A = 100.0\ \text{N}.$
At $t=2.00$: $150.0 = A + B(2.00)^2 = 100.0 + 4B$. So $4B = 50.0 \Rightarrow B = 12.5$. Units: $B$ multiplies $t^2$, so units are N/s².
Thus $A = 100.0\ \text{N},\; B = 12.5\ \text{N/s}^2.$

(b) Rocket mass $m = 8.00\ \text{kg}$. Weight $W = mg = 8.00(9.80) = 78.4\ \text{N}.$

(i) At $t=0^+$: $F = 100.0\ \text{N}.$ Net force up: $F_{\text{net}} = F - W = 100.0 - 78.4 = 21.6\ \text{N}.$
Acceleration $a = F_{\text{net}}/m = 21.6/8.00 = 2.700\ \text{m/s}^2.$

(ii) At $t=3.00\ \text{s}$: $F = 100.0 + 12.5(3.00)^2 = 100.0 + 12.5(9) = 100.0 + 112.5 = 212.5\ \text{N}.$
Net force: $F_{\text{net}} = 212.5 - 78.4 = 134.1\ \text{N}.$
Acceleration: $a = 134.1/8.00 = 16.7625\ \text{m/s}^2 \approx 16.76\ \text{m/s}^2.$

(c) Far from gravity: net force = applied force $F = 212.5\ \text{N}$. Acceleration $a = F/m = 212.5/8.00 = 26.5625\ \text{m/s}^2 \approx 26.56\ \text{m/s}^2.$

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Q4.16 Astronaut’s pack weighs $17.5\ \text{N}$ on Earth and $3.24\ \text{N}$ on a moon. (a) Find moon’s $g$. (b) Mass of pack.

Answer (stepwise):
(a) Mass $m = W_{\text{Earth}}/g_{\text{Earth}} = 17.5/9.80 = 1.7857142857\ \text{kg} \approx 1.7857\ \text{kg}.$
Moon’s $g = W_{\text{moon}}/m = 3.24 / 1.7857142857 = 1.8144\ \text{m/s}^2 \approx 1.81\ \text{m/s}^2.$

(b) Mass on the moon is the same mass: $m \approx 1.7857\ \text{kg}.$

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Q4.17 Superman throws a 2400-N boulder. What horizontal force gives it horizontal acceleration $12.0\ \text{m/s}^2$?

Answer (stepwise):
First get mass: $m = W/g = 2400/9.80 = 244.8979591837\ \text{kg} \approx 244.90\ \text{kg}.$
Force needed: $F = m a = 244.8979591837 \times 12.0 = 2938.7755102044\ \text{N} \approx 2.94\times10^3\ \text{N}.$

So $F \approx 2939\ \text{N}.$

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Q4.18 (a) Flea mass $=210\ \text{mg}$. (b) Froghopper mass $=12.3\ \text{mg}$. (c) House cat weight $=45\ \text{N}$. Compute weights/mass/convert to lb.

Answer (stepwise):
Convert units: $1\ \text{mg} = 1.0\times10^{-6}\ \text{kg}.$

(a) Flea mass $m = 210\times10^{-6} = 2.10\times10^{-4}\ \text{kg}.$ Weight $W = mg = 2.10\times10^{-4}\times9.80 = 2.058\times10^{-3}\ \text{N} \approx 2.06\times10^{-3}\ \text{N}.$

(b) Froghopper mass $m = 12.3\times10^{-6} = 1.23\times10^{-5}\ \text{kg}.$ Weight $W = 1.23\times10^{-5}\times9.80 = 1.2054\times10^{-4}\ \text{N} \approx 1.21\times10^{-4}\ \text{N}.$

(c) House cat: weight $W=45\ \text{N}.$ Pounds: $1\ \text{lbf}=4.44822\ \text{N}$. So mass in pounds $=45/4.44822 = 10.116\ \text{lb} \approx 10.12\ \text{lb}.$
Mass in kg $m = W/g = 45/9.80 = 4.5918367347\ \text{kg} \approx 4.59\ \text{kg}.$

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Q4.19 At Io, $g=1.81\ \text{m/s}^2$. Watermelon weighs $44.0\ \text{N}$ on Earth. (a) Mass on Earth. (b) Mass and weight on Io.

Answer (stepwise):
(a) Mass $m = 44.0/9.80 = 4.4897959184\ \text{kg} \approx 4.49\ \text{kg}.$
(b) Mass on Io is same $m \approx 4.49\ \text{kg}.$ Weight on Io $W_{\text{Io}} = m g_{\text{Io}} = 4.4897959184\times1.81 = 8.12653\ \text{N} \approx 8.13\ \text{N}.$

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Q4.20 Small car mass $380\ \text{kg}$ pushes truck mass $900\ \text{kg}$ east. Car exerts $1600\ \text{N}$ on truck. What magnitude does truck exert on car?

Answer: By Newton’s third law the truck exerts an equal and opposite force on the car. Magnitude $=1600\ \text{N}.$

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Q4.21 World-class sprinter: $m=55\ \text{kg}$, acceleration $=15\ \text{m/s}^2$. (a) How much horizontal force must sprinter exert on starting blocks? (b) Which body exerts the force that propels the sprinter: the blocks or the sprinter herself?

Answer (stepwise):
(a) Required horizontal force $F = m a = 55\times15 = 825\ \text{N}.$

(b) The blocks provide the reaction force that propels the sprinter forward: the sprinter pushes backward on the blocks (her muscles generate that push), and the blocks push forward on her (reaction), accelerating her. So the blocks exert the forward propulsive force on the sprinter (but the sprinter’s muscles are what create the push on the blocks in the first place).

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Find the rest question answers directly from manual 4.22-4.6

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