2 Marks .
A1. Define work in terms of force and displacement.
Work is done when a force causes a displacement of an object. It is defined as the product of the force component along the direction of displacement and the displacement.
A2. Differentiate positive and negative work.
-
Positive work: Done when force and displacement are in the same direction.
-
Negative work: Done when force and displacement are in opposite directions.
A3. How is work calculated when the force and displacement are not in the same direction?
Work is calculated using
[
W = Fd\cos\theta,
]
where θ is the angle between force and displacement.
A4. What is meant by kinetic energy?
Kinetic energy is the energy possessed by a body due to its motion.
[
K.E. = \frac{1}{2}mv^2
]
A5. How does the work-energy theorem connect total work and kinetic energy?
The work-energy theorem states that the total work done on an object equals the change in its kinetic energy.
A6. What is the physical significance of the work-energy theorem?
It shows that forces acting on a body change its motion by doing work, and this work directly converts into changes in kinetic energy.
A7. What is the difference between the kinetic energy of a composite system and a single particle system?
A composite system’s kinetic energy includes the kinetic energy of all particles (internal + motion of center of mass), while a single particle has only one kinetic energy term.
A8. If there is a net nonzero force on a moving object, can the total work done on the object be zero? Explain.
Yes. If the displacement is perpendicular to the force,
[
W = Fd\cos 90^\circ = 0.
]
Example: A body undergoing uniform circular motion—centripetal force is perpendicular to displacement, so work done is zero.
A9. State the principle of conservation of energy. Provide an example.
Energy can neither be created nor destroyed; it only transforms from one form to another.
Example: A falling object converts potential energy to kinetic energy.
A10. Define work in physics and state the formula used to calculate it.
Work is done when a force acting on a body causes displacement.
Formula:
[
W = Fd
]
(or (W = Fd\cos\theta) when not in same direction).
A11. If a force acts on an object at an angle to its displacement, how do you calculate the work done?
Use
[
W = Fd\cos\theta,
]
where θ is the angle between force and displacement.
A12. What is the SI unit of work? Express it in terms of SI units of force and distance.
The SI unit of work is the joule (J).
[
1 , \text{J} = 1, \text{N} \times 1, \text{m} = (kg, m/s^2) \times m.
]
A13. An object of mass 2 kg is lifted 2 m vertically. Calculate the work done.
[
W = mgh = 2 \times 9.8 \times 2 = 39.2, \text{J}
]
(≈ 40 J if (g = 10 , m/s^2)).
A14. What is total work, and how is it calculated for multiple forces?
Total work is the sum of the work done by all forces acting on an object.
[
W_{\text{total}} = \sum W_i
]
A15. State the work-energy theorem.
The net work done on an object is equal to the change in its kinetic energy:
[
W_{\text{net}} = \Delta K.
]
A16. How does the kinetic energy of a body change if its speed doubles?
Kinetic energy depends on (v^2). If speed doubles:
[
K.E. \propto v^2 \Rightarrow (2v)^2 = 4v^2
]
So, kinetic energy becomes four times.
6 Marks .
Below are clear, step-by-step solutions for all problems B1–B11.
B1. Work done by each force
Given:
Distance (d = 20\text{ m})
Weight = 14,700 N (but no vertical motion → gravity & normal do zero work)
Applied force = (5000\text{ N}) at (36.9^\circ)
Friction = (3500\text{ N}) opposite motion
(a) Work done by tractor force
[
W_{F} = F d \cos\theta
= 5000(20)(\cos 36.9^\circ)
]
[
\cos 36.9^\circ \approx 0.8
]
[
W_F = 5000 \times 20 \times 0.8 = 80,000\text{ J}
]
(b) Work done by friction
[
W_f = - f d = -3500(20) = -70,000\text{ J}
]
(c) Work done by gravity and normal
No vertical displacement:
[
W_g = 0,\qquad W_N = 0
]
Total work
[
W_{\text{total}} = 80,000 - 70,000 = 10,000\text{ J}
]
B2. Work on the book
Given:
Push = 2.40 N
Friction = 0.600 N
Displacement = 1.50 m
(a) Work by you
[
W_{\text{push}} = Fd = 2.40(1.50) = 3.60\text{ J}
]
(b) Work by friction
[
W_f = -0.600(1.50) = -0.900\text{ J}
]
(c) Work by normal force
Perpendicular to motion → 0 J
(d) Work by gravity
Vertical displacement = 0 → 0 J
(e) Net work
[
W_{\text{net}} = 3.60 - 0.90 = 2.70\text{ J}
]
B3. Work and speed between points A, B, C
Mass (m = 1.50\text{ kg})
(a) Work from A → B
[
W = \Delta K = \tfrac12 m(v_B^2 - v_A^2)
]
[
W = \tfrac12(1.50)(1.25^2 - 3.21^2)
]
[
= 0.75(1.5625 - 10.3041) = -6.55\text{ J}
]
(b) Work −0.750 J from B → C
[
\Delta K = -0.750
]
[
K_C = K_B - 0.750
]
[
\tfrac12 m v_C^2 = \tfrac12(1.5)(1.25^2) - 0.750
]
[
= 1.1719 - 0.750 = 0.4219
]
[
v_C = \sqrt{\frac{2(0.4219)}{1.5}} = 0.75\text{ m/s}
]
(c) If +0.750 J done on book
[
K_C = 1.1719 + 0.750 = 1.9219
]
[
v_C = \sqrt{\frac{2(1.9219)}{1.5}} = 1.60\text{ m/s}
]
B4. Throwing a rock
Weight = 3.00 N → mass
[
m = \frac{3}{9.8} = 0.306\text{ kg}
]
At height 15 m, speed = 25 m/s upward.
(a) Initial speed at ground
Apply W–E from ground → 15 m:
[
W_g = -mgh = -3.00(15) = -45\text{ J}
]
[
\tfrac12 m v_0^2 + W_g = \tfrac12 m v^2
]
[
\tfrac12 m v_0^2 - 45 = \tfrac12 m (25^2)
]
[
\tfrac12 m v_0^2 = \tfrac12 m(625)+45
]
Divide both sides by (\tfrac12 m):
[
v_0^2 = 625 + \frac{90}{0.306} = 625 + 294 = 919
]
[
v_0 = 30.3\text{ m/s}
]
(b) Maximum height
At top → final speed = 0:
[
mgh_{\max} = \tfrac12 m v_0^2
]
[
h_{\max} = \frac{v_0^2}{2g}
= \frac{(30.3)^2}{19.6}
= 46.9\text{ m}
]
B5. Work and angle dependence
Work:
Work is the energy transferred when a force causes displacement.
If force ( \vec F ) acts through displacement ( \vec d ):
[
W = \vec F \cdot \vec d = Fd\cos\theta
]
Where θ is angle between force and displacement.
-
If (0^\circ): maximum work
-
If (90^\circ): zero work
-
If (180^\circ): negative work
B6. Work–Energy Theorem & KE derivation
Statement:
The net work done on an object equals its change in kinetic energy.
[
W_{\text{net}} = \Delta K
]
Starting with Newton’s second law:
[
F = ma
]
Work done by constant force:
[
W = Fd = ma d
]
Using kinematics:
[
v^2 = v_0^2 + 2ad \Rightarrow ad = \frac{v^2 - v_0^2}{2}
]
[
W = m \frac{v^2 - v_0^2}{2}
]
[
W = \tfrac12 mv^2 - \tfrac12 mv_0^2
]
Thus kinetic energy:
[
K = \tfrac12 mv^2
]
B7. Work on block on incline
Mass = 5 kg (not needed)
Horizontal force = 20 N
Distance along incline = 4 m
Angle incline = 30°
Horizontal force component along incline:
[
F_{\parallel} = F\cos 30^\circ = 20(0.866) = 17.32\text{ N}
]
Work done
[
W = F_{\parallel} d = 17.32(4) = 69.3\text{ J}
]
On frictionless incline all work increases KE:
[
\Delta K = 69.3\text{ J}
]
B8. Pile-driver hammer
Mass = 200 kg
Drop height = 3 m
Friction during fall = 60 N
Beam driven distance = 7.4 cm = 0.074 m
(a) Speed before impact
Work done by gravity:
[
W_g = mgh = 200(9.8)(3) = 5880\text{ J}
]
Work done by friction:
[
W_f = -60(3) = -180\text{ J}
]
Net work:
[
W = 5700\text{ J}
]
Using (W = \tfrac12 mv^2):
[
v = \sqrt{\frac{2W}{m}} = \sqrt{\frac{11400}{200}} = 7.55\text{ m/s}
]
(b) Average force exerted on I-beam
Kinetic energy at impact:
[
K = 5700\text{ J}
]
This energy is used to push beam down 0.074 m:
[
F_{\text{avg}} = \frac{K}{d} = \frac{5700}{0.074} = 77,000\text{ N}
]
B9. Energy consumption
Power = 1.5 kW
Time = 8 h
Energy per day
[
E = Pt = 1.5 \times 8 = 12\text{ kWh}
]
Cost for 30 days
Daily cost:
[
12 \times 6 = 72\text{ Rs}
]
Monthly cost:
[
72 \times 30 = 2160\text{ Rs}
]
B10. Two ice boats, masses m and 2m
Both experience same constant force (F) and same displacement (s).
Work done:
[
W = Fs
]
Since the same work is done on both, both gain the same kinetic energy.
Thus both boats cross the finish line with equal kinetic energy,
even though the lighter boat reaches a higher speed.
B11. Force on sled
Mass = 12 kg
Initial speed = 4 m/s
Final speed = 6 m/s
Distance = 2.50 m
Use W–E:
[
W = \Delta K = \tfrac12 m (v^2 - v_0^2)
]
[
W = \tfrac12(12)(36 - 16) = 6(20) = 120\text{ J}
]
Work by constant force:
[
W = Fd \Rightarrow F = \frac{120}{2.5} = 48\text{ N}
]
14 Marks .
Below are complete, well-structured, exam-ready answers for C1–C6.
C1 (7 marks)
Define work. Explain positive, negative, and zero work with examples.
Definition of Work:
Work is defined as the energy transferred to or from an object when a force acting on it causes displacement. For a constant force,
[
W = Fd\cos\theta
]
where
-
(F) = force applied,
-
(d) = displacement,
-
(\theta) = angle between force and displacement.
1. Positive Work
Positive work is done when the force and displacement are in the same direction. The object’s kinetic energy increases.
Example:
-
A person pushing a box forward.
-
The applied force and displacement are both forward, so ( \cos 0^\circ = 1).
-
Work done is positive.
2. Negative Work
Negative work is done when the force acts opposite to displacement. It reduces an object’s kinetic energy.
Example:
-
Friction acting on a moving object.
-
A moving car slows down because friction and air resistance oppose its motion.
3. Zero Work
Zero work occurs when the force is perpendicular to displacement or when there is no displacement.
Examples:
-
A person carrying a load horizontally: force is vertical, displacement is horizontal → ( \theta = 90^\circ), so work = 0.
-
A book resting on a table: no displacement → work = 0.
C2 (14 marks)
Explain the work–energy theorem for straight-line and curved motion with varying forces.
Work–Energy Theorem (Statement)
The net work done by all forces on a particle equals the change in its kinetic energy:
[
W_{\text{net}} = \Delta K = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2
]
A. Motion Along a Straight Line (1D Motion)
If forces act along a straight line, the work done by the net force is:
[
W_{\text{net}} = \int_{x_1}^{x_2} F_{\text{net}}(x), dx
]
This applies whether the force is constant or variable.
Explanation:
-
When the net force accelerates the object, kinetic energy increases.
-
When the net force opposes motion, kinetic energy decreases.
Example:
A car accelerating on a straight road:
Engine force does positive work, friction does negative work. Net work = gain in kinetic energy.
B. Motion Along a Curved Path
For motion along any curved trajectory, work done is:
[
W_{\text{net}} = \int \vec{F} \cdot d\vec{r}
]
Here we consider the component of force tangent to the path.
Key points:
-
Only the tangential component of force changes the speed.
-
Normal force (perpendicular to displacement) does zero work.
-
Even along curved paths, the theorem still holds:
[
\Delta K = W_{\text{net}}
]
Example:
Object sliding down a curved slide:
Gravity has a component along the slide that speeds up the object; normal force does zero work.
Conclusion:
The work–energy theorem applies universally to both straight and curved paths and for any varying force. It is a powerful tool for analyzing how forces change an object’s speed.
C3 (14 marks)
Explain power, instantaneous power, average power, and relation with velocity.
Power (Definition):
Power is the rate at which work is done or energy is transferred.
[
P = \frac{W}{t}
]
1. Average Power
Average power is total work done divided by total time:
[
P_{\text{avg}} = \frac{W_{\text{total}}}{\Delta t}
]
Example:
A crane lifting a load in 10 seconds vs. 5 seconds. The one that does it faster has higher power.
2. Instantaneous Power
Instantaneous power is power at an exact moment:
[
P = \vec{F} \cdot \vec{v}
]
This is important when the force or velocity changes continuously (e.g., a car during acceleration).
3. Relation Between Power and Velocity
Since:
[
P = Fv\cos\theta
]
-
If force and velocity are in the same direction → (P = Fv).
-
Higher velocity → higher power required to keep force acting.
Example:
A car engine must deliver higher power at high speeds because resistive forces increase and velocity increases.
Significance of Power
-
Determines how quickly energy is used.
-
Important in designing engines, motors, and electrical appliances.
C4 (14 marks)
Graph of (F) vs. (x) such that work = 0 from (x_1) to (x_2), but force ≠ 0
Required condition:
Work is the area under the curve:
[
W = \int_{x_1}^{x_2} F(x), dx
]
Work = 0 if positive area = negative area.
Sketch Description
A suitable graph is:
-
Force is positive for part of the interval.
-
Force is negative for the remaining part.
-
Areas above and below the x-axis cancel out.
Example graph:
A sine or wave-like curve crossing the x-axis, with equal positive and negative lobes.
Physical Example:
A mass attached to a spring:
-
When the spring is compressed, force is negative (restoring).
-
When stretched, force is positive.
-
Over one complete oscillation from (x_1) to (x_2), net work is zero.
This is because the spring stores and returns energy, resulting in zero net work though force is always present.
C5 (14 marks)
Discuss the work–energy theorem, significance, real-life applications, and relation to conservation of energy.
Work–Energy Theorem:
The net work done on an object equals the change in its kinetic energy:
[
W_{\text{net}} = \Delta K
]
Significance:
-
Simplifies dynamics by avoiding direct use of Newton’s laws.
-
Allows analysis of systems with varying forces.
-
Connects force (a mechanical concept) with energy (a more general concept).
-
Forms the foundation of energy conservation.
Interrelation of Work and Energy
-
Work transfers energy.
-
Kinetic energy increases when positive work is done.
-
Potential energy changes when conservative forces do work.
-
The theorem explains how forces change motion.
Real-Life Applications
1. Driving a Car
-
Engine does positive work → car speeds up.
-
Brakes do negative work → kinetic energy is removed as heat.
2. Lifting an Object
-
Work done by lifting force increases gravitational potential energy.
-
When the object is lowered, gravity does positive work.
3. Sports
-
A baseball bat does work on the ball, increasing its kinetic energy.
4. Roller Coasters
-
Work done by motors at the start is converted into potential energy.
-
As the coaster descends, gravity does work and speeds it up.
Importance in Physics
-
Core principle in mechanics.
-
Helps analyze systems where forces change with position.
-
Leads directly to conservation of mechanical energy for conservative forces.
C6 (14 marks)
Explain work done by a variable force and how to calculate it.
Variable Force
When a force changes with displacement, the simple formula (W = Fd\cos\theta) cannot be used.
Instead, we calculate work by integration:
[
W = \int_{x_1}^{x_2} F(x), dx
]
This gives the area under the force–displacement graph.
Example: Stretching a Spring
A spring obeys Hooke’s Law:
[
F = kx
]
Work done to stretch it from 0 to (x):
[
W = \int_0^x kx,dx = \frac{1}{2}kx^2
]
This work is stored as elastic potential energy.
Practical Applications
1. Stretching a Spring
Used in:
-
Clocks,
-
Toys,
-
Vehicle suspension systems.
The force increases as the spring stretches, so we must calculate work using integration.
2. Pulling an Object with Changing Resistance
Example: pulling a rope out of water, where the weight of the rope decreases as more rope leaves the water.
Force varies → work must be computed with integration.
3. Charging a Bow (Archery)
The force to pull a bowstring increases nonlinearly.
Work = area under the force–stretch curve.
Summary
-
Variable forces require calculus.
-
Work = area under the F–x graph.
-
Many real systems involve variable forces, making this concept essential in engineering and physics.