2 Mark
A1. What is kinetic energy and how is it related to work done?
Kinetic energy is the energy an object has due to its motion.
[
K = \tfrac12 mv^2
]
The work done on an object changes its kinetic energy, as stated by the work–energy theorem:
[
W_{\text{net}} = \Delta K
]
A2. What is elastic potential energy? Give its formula.
Elastic potential energy is the energy stored in an elastic body (like a stretched or compressed spring).
[
U = \tfrac12 kx^2
]
A3. Explain how gravitational potential energy changes when a basketball is thrown upwards.
As the basketball rises, its height increases, so gravitational potential energy increases. This increase comes from a decrease in its kinetic energy (the ball slows down).
A4. What is potential energy?
Potential energy is the energy an object possesses due to its position or configuration in a force field (e.g., gravitational or elastic).
A5. Define kinetic energy.
Kinetic energy is the energy of motion, given by:
[
K = \tfrac12 mv^2
]
A6. State the law of conservation of energy.
Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system remains constant.
A7. Does gravitational potential energy depend on straight or curved path? Explain.
No. Gravitational potential energy depends only on vertical height, not on the path taken. A straight or curved path reaching the same height results in the same potential energy.
A8. Explain the relationship between work and kinetic energy.
The net work done on an object changes its kinetic energy.
[
W_{\text{net}} = \Delta K
]
Positive work increases kinetic energy; negative work decreases it.
A9. What is gravitational potential energy? Give its formula.
It is the energy an object has due to its height in a gravitational field.
[
U = mgh
]
A10. How is potential energy converted to kinetic energy during free fall?
As an object falls, its height decreases, so gravitational potential energy decreases. This lost potential energy becomes kinetic energy, increasing the object’s speed.
A11. What is an elastic body? Provide an example.
An elastic body returns to its original shape after the deforming force is removed.
Example: A spring or a rubber band.
A12. Describe energy transformations when a spring is stretched and released.
-
When stretched: work done on the spring becomes elastic potential energy.
-
When released: that elastic potential energy converts into kinetic energy of the moving spring/mass.
A13. What does "elastic potential energy" mean in a spring system?
It refers to the energy stored in a spring due to its deformation (stretching or compression).
A14. How do you calculate the work done by gravitational force?
Work done by gravity during vertical displacement (h) is:
[
W = mgh
]
Positive when moving downward, negative when moving upward.
A15. What is total mechanical energy, and how is it conserved?
Total mechanical energy is the sum of kinetic and potential energies:
[
E = K + U
]
In an isolated system with no non-conservative forces (friction), total mechanical energy remains constant.
6 Mark
Below are complete, step-by-step, 6-mark level answers for B1–B6 (clear, well-explained, exam-ready).
B1 (6 Marks)
Prove that work done by gravity equals the change in gravitational potential energy.
Consider an object of mass (m) lifted vertically upward by height (h).
Gravitational force acts downward:
[
F_g = mg
]
Displacement upward:
[
\vec{d} = h \text{ (up)}
]
Angle between force and displacement:
[
\theta = 180^\circ
]
Thus, work done by gravity:
[
W_g = F_g d \cos \theta = mgh (-1)
]
[
\boxed{W_g = -mgh}
]
Gravitational potential energy is defined as:
[
U = mgh
]
Change in potential energy:
[
\Delta U = U_f - U_i = mgh - 0 = mgh
]
Comparing:
[
W_g = -\Delta U
]
Thus, work done by gravity is equal to the negative change in gravitational potential energy.
Calculate maximum height for a ball thrown up with (u = 15\ \text{m/s})
At maximum height, final velocity = 0.
Using energy method:
[
\tfrac12 m u^2 = mgh
]
[
h = \frac{u^2}{2g} = \frac{15^2}{2 \times 9.8}
]
[
h = \frac{225}{19.6} = 11.48\text{ m}
]
[
\boxed{h_{\max} = 11.5\text{ m}}
]
B2 (6 Marks)
Prove that work done by elastic force equals change in elastic potential energy.
Spring force:
[
F = -kx
]
Work done from position (x) to 0:
[
W = \int_x^0 (-kx), dx
]
[
W = -\left[\frac12 kx^2\right]_x^0 = \frac12 kx^2
]
Elastic potential energy stored:
[
U = \tfrac12 kx^2
]
Thus,
[
\boxed{W = \Delta U}
]
Work done by spring equals decrease in elastic potential energy.
Numerical: Work done by a spring (k = 200\text{ N/m}), compressed (0.3\text{ m})
[
W = \tfrac12 kx^2
]
[
W = \tfrac12 (200)(0.3)^2 = 100(0.09)
]
[
W = 9\text{ J}
]
[
\boxed{W = 9\text{ J}}
]
B3 (6 Marks)
Law of Conservation of Energy
Energy cannot be created or destroyed; it changes from one form to another.
Total energy of an isolated system remains constant.
Proof:
Let the system undergo changes in:
-
Kinetic energy (K)
-
Potential energy (U)
-
Internal energy (E_{\text{int}})
Net work done on the system:
[
W_{\text{ext}} = \Delta K
]
Work done by conservative forces:
[
W_c = -\Delta U
]
Work done by non-conservative forces (friction etc.) converts to internal energy:
[
W_{nc} = \Delta E_{\text{int}}
]
Total work:
[
W_{\text{ext}} = W_c + W_{nc}
]
[
\Delta K = -\Delta U + \Delta E_{\text{int}}
]
Rearrange:
[
\Delta K + \Delta U + \Delta E_{\text{int}} = 0
]
Thus:
[
\boxed{\Delta K + \Delta U + \Delta E_{\text{int}} = 0}
]
Total energy remains constant.
B4 (6 Marks)
Given:
Elevator mass (m = 2000\text{ kg})
Initial speed (v = 4\text{ m/s})
Spring compression (x = 2.00\text{ m})
Safety clamp force (F_c = 17,000\text{ N}) upward
Energy at moment of contact:
Initial kinetic energy:
[
K_i = \frac12 mv^2 = \frac12(2000)(4^2) = 16,000\text{ J}
]
Work done by gravity during 2 m fall:
[
W_g = mgh = (2000)(9.8)(2) = 39,200\text{ J}
]
Work done by clamp (opposes motion):
[
W_c = -F_c x = -17000(2) = -34,000\text{ J}
]
Total energy to be absorbed by the spring:
[
E_{\text{spring}} = K_i + W_g + W_c
]
[
E_{\text{spring}} = 16000 + 39200 - 34000
]
[
E_{\text{spring}} = 21,200\text{ J}
]
Spring energy stored:
[
E_{\text{spring}} = \frac12 kx^2
]
[
21200 = \frac12 k (2)^2 = 2k
]
[
k = 10,600\ \text{N/m}
]
[
\boxed{k = 1.06 \times 10^4\ \text{N/m}}
]
B5 (6 Marks)
Mass = 25 kg
Quarter circle radius (R = 3\text{ m})
(a) Speed at bottom
Height drop:
[
h = R = 3\text{ m}
]
Using energy conservation:
[
mgh = \tfrac12 mv^2
]
[
v = \sqrt{2gR}
]
[
v = \sqrt{2(9.8)(3)} = \sqrt{58.8} = 7.67\text{ m/s}
]
[
\boxed{v = 7.67\text{ m/s}}
]
(b) Normal force at bottom
At bottom of circular path:
[
N - mg = \frac{mv^2}{R}
]
[
N = mg + \frac{mv^2}{R}
]
[
N = 25(9.8) + \frac{25(7.67^2)}{3}
]
[
N = 245 + \frac{25(58.8)}{3}
]
[
N = 245 + 490 = 735\text{ N}
]
[
\boxed{N = 735\text{ N}}
]
B6 (6 Marks)
Mass (m = 200\text{ g} = 0.20\text{ kg})
Spring constant (k = 500\text{ N/m})
Initial stretch = 0.10 m
Find velocity at (x = 0.08\text{ m})
Use energy conservation
Initial total energy:
[
E_i = \tfrac12 k(0.10)^2 = \tfrac12 (500)(0.01)= 2.5\text{ J}
]
Elastic energy at (x=0.08):
[
U = \tfrac12 k(0.08)^2 = 250(0.0064)=1.6\text{ J}
]
Kinetic energy at 0.08 m:
[
K = E_i - U = 2.5 - 1.6 = 0.9\text{ J}
]
Velocity:
[
\tfrac12 mv^2 = 0.9
]
[
v = \sqrt{\frac{2(0.9)}{0.20}} = \sqrt{9} = 3\text{ m/s}
]
[
\boxed{v = 3.0\text{ m/s}}
]
If you want, I can prepare A, B, C type combined notes, or a PDF-like final notes version.