Work Energy and Power Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 2.5 \, \text{kg} \) object is dropped from \( 13 \, \text{m} \). What is its kinetic energy just before hitting the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( V = mgh = 2.5 \times 10 \times 13 = 325 \, \text{J} \).

By conservation, \( K_f = V_i = 325 \, \text{J} \).

300 J
325 J
350 J
375 J
2

A \( 6 \, \text{kg} \) block is pushed with \( 30 \, \text{N} \) over \( 5 \, \text{m} \) on a frictionless surface, starting from rest. What is its final speed?

Work \( W = 30 \times 5 = 150 \, \text{J} \).

By work-energy theorem, \( K_f = W \Rightarrow \frac{1}{2} \times 6 \times v^2 = 150 \Rightarrow v^2 = 50 \Rightarrow v = \sqrt{50} \approx 7.07 \, \text{m/s} \).

6.5 m/s
7 m/s
7.5 m/s
8 m/s
2

A person applies a force of \( 120 \, \text{N} \) over \( 5 \, \text{m} \), then the force decreases linearly to \( 40 \, \text{N} \) over \( 10 \, \text{m} \). What is the total work done?

Work in first part: \( W_1 = 120 \times 5 = 600 \, \text{J} \).

Work in second part: \( W_2 = \frac{1}{2} (120 + 40) \times 10 = 80 \times 10 = 800 \, \text{J} \).

Total work: \( W = 600 + 800 = 1400 \, \text{J} \).

1300 J
1400 J
1500 J
1600 J
2

A \( 1600 \, \text{kg} \) car at \( 15 \, \text{m/s} \) hits a spring (\( k = 10^4 \, \text{N/m} \)). What is the maximum compression?

Initial \( K = \frac{1}{2} \times 1600 \times 15^2 = 180000 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 180000 \Rightarrow 5000 x_m^2 = 180000 \Rightarrow x_m = \sqrt{36} = 6 \, \text{m} \).

5.8 m
6 m
6.2 m
6.5 m
2

A \( 3 \, \text{kg} \) block slides down a frictionless incline from \( 7 \, \text{m} \) height. What is its speed at the bottom? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( mgh = 3 \times 10 \times 7 = 210 \, \text{J} \).

Kinetic energy \( \frac{1}{2} m v^2 = 210 \Rightarrow v^2 = 140 \Rightarrow v = \sqrt{140} \approx 11.83 \, \text{m/s} \).

11 m/s
11.8 m/s
12.5 m/s
13 m/s
2

A \( 0.4 \, \text{kg} \) pendulum bob completes a vertical circle of radius \( 1.8 \, \text{m} \). What is the speed at the bottom? (Take \( g = 10 \, \text{m/s}^2 \))

At top, \( v_C = \sqrt{gL} \), energy conservation: \( \frac{1}{2} m v_0^2 = \frac{1}{2} m (gL) + 2mgL \).

\( v_0^2 = 5gL = 5 \times 10 \times 1.8 = 90 \Rightarrow v_0 = \sqrt{90} \approx 9.49 \, \text{m/s} \).

9 m/s
9.5 m/s
10 m/s
10.5 m/s
2

A \( 6 \, \text{kg} \) mass falls from \( 4 \, \text{m} \) onto a spring (\( k = 1500 \, \text{N/m} \)). What is the maximum compression? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( mgh = 6 \times 10 \times 4 = 240 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 240 \Rightarrow 750 x_m^2 = 240 \Rightarrow x_m = \sqrt{0.32} \approx 0.566 \, \text{m} \).

0.5 m
0.57 m
0.6 m
0.65 m
2

A \( 0.8 \, \text{kg} \) pendulum bob completes a vertical circle of radius \( 2.5 \, \text{m} \). What is the speed at the top? (Take \( g = 10 \, \text{m/s}^2 \))

At top, minimum speed \( v_C = \sqrt{gL} = \sqrt{10 \times 2.5} = \sqrt{25} = 5 \, \text{m/s} \).

4.5 m/s
5 m/s
5.5 m/s
6 m/s
2

A \( 5 \, \text{kg} \) mass at \( 8 \, \text{m/s} \) collides inelastically with a stationary \( 3 \, \text{kg} \) mass. What is the final speed?

Momentum conservation: \( 5 \times 8 = (5 + 3) v_f \Rightarrow 40 = 8 v_f \Rightarrow v_f = 5 \, \text{m/s} \).

4 m/s
5 m/s
6 m/s
7 m/s
2

A \( 4 \, \text{kg} \) mass moving at \( 5 \, \text{m/s} \) collides inelastically with a stationary \( 6 \, \text{kg} \) mass. What is their common speed after collision?

Momentum conservation: \( 4 \times 5 = (4 + 6) v_f \Rightarrow 20 = 10 v_f \Rightarrow v_f = 2 \, \text{m/s} \).

1.5 m/s
2 m/s
2.5 m/s
3 m/s
2

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