Work Energy and Power Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A force starts at \( 50 \, \text{N} \) over \( 6 \, \text{m} \), then decreases linearly to \( 10 \, \text{N} \) over \( 4 \, \text{m} \). What is the total work done?

Work in first part: \( W_1 = 50 \times 6 = 300 \, \text{J} \).

Work in second part: \( W_2 = \frac{1}{2} (50 + 10) \times 4 = 30 \times 4 = 120 \, \text{J} \).

Total work: \( W = 300 + 120 = 420 \, \text{J} \).

400 J
420 J
440 J
460 J
2

A \( 10 \, \text{kg} \) block is pushed with \( 50 \, \text{N} \) over \( 8 \, \text{m} \) on a frictionless surface, starting from \( 1 \, \text{m/s} \). What is its final speed?

Initial \( K_i = \frac{1}{2} \times 10 \times 1^2 = 5 \, \text{J} \).

Work \( W = 50 \times 8 = 400 \, \text{J} \).

Final \( K_f = K_i + W = 5 + 400 = 405 \, \text{J} \Rightarrow v_f = \sqrt{\frac{2 \times 405}{10}} = \sqrt{81} = 9 \, \text{m/s} \).

8 m/s
9 m/s
10 m/s
11 m/s
2

A spring (\( k = 300 \, \text{N/m} \)) is compressed by \( 0.2 \, \text{m} \). What is the potential energy stored?

Potential energy \( V = \frac{1}{2} k x^2 = \frac{1}{2} \times 300 \times (0.2)^2 = 150 \times 0.04 = 6 \, \text{J} \).

5 J
6 J
7 J
8 J
2

A \( 12 \, \text{kg} \) block is pushed with \( 60 \, \text{N} \) over \( 5 \, \text{m} \) on a frictionless surface, starting from rest. What is its final speed?

Work \( W = 60 \times 5 = 300 \, \text{J} \).

By work-energy theorem, \( K_f = W \Rightarrow \frac{1}{2} \times 12 \times v^2 = 300 \Rightarrow v^2 = 50 \Rightarrow v = \sqrt{50} \approx 7.07 \, \text{m/s} \).

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

A \( 0.9 \, \text{g} \) drop falls from \( 150 \, \text{m} \) and hits the ground at \( 13 \, \text{m/s} \). What is the work done by air resistance? (Take \( g = 10 \, \text{m/s}^2 \))

Work by gravity \( W_g = mgh = 0.0009 \times 10 \times 150 = 1.35 \, \text{J} \).

Final \( K = \frac{1}{2} \times 0.0009 \times 13^2 = 0.07605 \, \text{J} \).

\( K_f = W_g + W_r \Rightarrow 0.07605 = 1.35 + W_r \Rightarrow W_r = -1.27395 \, \text{J} \approx -1.27 \, \text{J} \).

-1.4 J
-1.27 J
-1.2 J
-1.1 J
2

A \( 10 \, \text{kg} \) mass is lifted \( 8 \, \text{m} \) in \( 5 \, \text{s} \) at constant speed. What is the power delivered? (Take \( g = 10 \, \text{m/s}^2 \))

Work \( W = mgh = 10 \times 10 \times 8 = 800 \, \text{J} \).

Power \( P = \frac{W}{t} = \frac{800}{5} = 160 \, \text{W} \).

140 W
160 W
180 W
200 W
2

A motor lifts a \( 2000 \, \text{kg} \) load at \( 3.5 \, \text{m/s} \) against \( 4000 \, \text{N} \) friction. What is the power? (Take \( g = 10 \, \text{m/s}^2 \))

Force \( F = mg + F_f = 2000 \times 10 + 4000 = 24000 \, \text{N} \).

Power \( P = F \cdot v = 24000 \times 3.5 = 84000 \, \text{W} \).

80 kW
84 kW
88 kW
92 kW
2

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

At top, minimum speed \( v_C = \sqrt{gL} = \sqrt{10 \times 2} = \sqrt{20} \approx 4.47 \, \text{m/s} \).

4 m/s
4.5 m/s
5 m/s
5.5 m/s
2

A \( 1 \, \text{g} \) drop falls from \( 800 \, \text{m} \) and hits the ground at \( 30 \, \text{m/s} \). What is the work done by air resistance? (Take \( g = 10 \, \text{m/s}^2 \))

Work by gravity \( W_g = mgh = 0.001 \times 10 \times 800 = 8 \, \text{J} \).

Final \( K = \frac{1}{2} \times 0.001 \times 30^2 = 0.45 \, \text{J} \).

By work-energy theorem, \( K_f = W_g + W_r \Rightarrow 0.45 = 8 + W_r \Rightarrow W_r = -7.55 \, \text{J} \).

-8 J
-7.55 J
-7 J
-6.5 J
2

A \( 45 \, \text{g} \) bullet at \( 1100 \, \text{m/s} \) emerges from a block with \( 50\% \) of its initial kinetic energy. What is its emergent speed?

Initial \( K = \frac{1}{2} \times 0.045 \times 1100^2 = 27225 \, \text{J} \).

Final \( K = 0.5 \times 27225 = 13612.5 \, \text{J} \).

\( \frac{1}{2} \times 0.045 \times v_f^2 = 13612.5 \Rightarrow v_f^2 = 605000 \Rightarrow v_f = \sqrt{605000} \approx 777.8 \, \text{m/s} \).

760 m/s
778 m/s
800 m/s
820 m/s
2

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