Work Energy and Power Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 25 \, \text{g} \) bullet at \( 600 \, \text{m/s} \) emerges from a target with \( 50\% \) of its initial kinetic energy. What is its emergent speed?

Initial \( K = \frac{1}{2} \times 0.025 \times 600^2 = 4500 \, \text{J} \).

Final \( K = 0.5 \times 4500 = 2250 \, \text{J} \).

\( \frac{1}{2} \times 0.025 \times v_f^2 = 2250 \Rightarrow v_f^2 = 180000 \Rightarrow v_f = \sqrt{180000} \approx 424.3 \, \text{m/s} \).

400 m/s
424 m/s
450 m/s
475 m/s
2

A \( 1100 \, \text{kg} \) car at \( 16 \, \text{m/s} \) hits a spring (\( k = 7 \times 10^3 \, \text{N/m} \)). What is the maximum compression?

Initial \( K = \frac{1}{2} \times 1100 \times 16^2 = 140800 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 140800 \Rightarrow 3500 x_m^2 = 140800 \Rightarrow x_m = \sqrt{40.228} \approx 6.34 \, \text{m} \).

6.1 m
6.3 m
6.5 m
6.7 m
2

A \( 2 \, \text{kg} \) block falls from \( 5 \, \text{m} \) onto a spring (\( k = 500 \, \text{N/m} \)). What is the maximum compression? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( mgh = 2 \times 10 \times 5 = 100 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 100 \Rightarrow 250 x_m^2 = 100 \Rightarrow x_m = \sqrt{0.4} \approx 0.63 \, \text{m} \).

0.5 m
0.6 m
0.63 m
0.7 m
3

A spring (\( k = 250 \, \text{N/m} \)) is compressed by \( 0.4 \, \text{m} \). What is the stored potential energy?

Potential energy \( V = \frac{1}{2} k x^2 = \frac{1}{2} \times 250 \times (0.4)^2 = 125 \times 0.16 = 20 \, \text{J} \).

18 J
20 J
22 J
24 J
2

A \( 7 \, \text{kg} \) block at \( 12 \, \text{m/s} \) encounters a force \( F = \frac{-2.4}{x} \, \text{N} \) from \( x = 1 \, \text{m} \) to \( x = 5 \, \text{m} \). What is the final speed?

Initial \( K = \frac{1}{2} \times 7 \times 12^2 = 504 \, \text{J} \).

Work \( W = \int_{1}^{5} \frac{-2.4}{x} \, dx = -2.4 \ln(5/1) = -2.4 \ln 5 \approx -3.86 \, \text{J} \).

Final \( K = 504 - 3.86 = 500.14 \, \text{J} \Rightarrow v_f = \sqrt{\frac{2 \times 500.14}{7}} \approx 11.95 \, \text{m/s} \).

11.7 m/s
12 m/s
12.2 m/s
12.5 m/s
2

A \( 1.2 \, \text{kg} \) pendulum bob completes a vertical circle of radius \( 1.6 \, \text{m} \). What is the speed at the top? (Take \( g = 10 \, \text{m/s}^2 \))

At top, minimum speed \( v_C = \sqrt{gL} = \sqrt{10 \times 1.6} = \sqrt{16} = 4 \, \text{m/s} \).

3.5 m/s
4 m/s
4.5 m/s
5 m/s
2

A spring (\( k = 400 \, \text{N/m} \)) is compressed from \( 0.2 \, \text{m} \) to \( 0.35 \, \text{m} \). What is the work done by the spring force?

Work \( W_s = \frac{1}{2} k (x_i^2 - x_f^2) = \frac{1}{2} \times 400 \times (0.2^2 - 0.35^2) = 200 \times (0.04 - 0.1225) = -16.5 \, \text{J} \).

-18 J
-16.5 J
-15 J
-13.5 J
2

A \( 20 \, \text{g} \) bullet at \( 1000 \, \text{m/s} \) emerges from a block with \( 70\% \) of its initial kinetic energy. What is its emergent speed?

Initial \( K = \frac{1}{2} \times 0.02 \times 1000^2 = 10000 \, \text{J} \).

Final \( K = 0.7 \times 10000 = 7000 \, \text{J} \).

\( \frac{1}{2} \times 0.02 \times v_f^2 = 7000 \Rightarrow v_f^2 = 700000 \Rightarrow v_f = \sqrt{700000} \approx 836.7 \, \text{m/s} \).

820 m/s
837 m/s
850 m/s
870 m/s
2

A ball of mass \( 0.5 \, \text{kg} \) is dropped from a height of \( 8 \, \text{m} \). What is its kinetic energy just before hitting the ground? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy at top: \( V = mgh = 0.5 \times 10 \times 8 = 40 \, \text{J} \).

By conservation of energy, \( K_f = V_i = 40 \, \text{J} \).

30 J
35 J
40 J
45 J
3

A force starts at \( 90 \, \text{N} \) over \( 3 \, \text{m} \), then decreases linearly to \( 40 \, \text{N} \) over \( 7 \, \text{m} \). What is the total work done?

Work in first part: \( W_1 = 90 \times 3 = 270 \, \text{J} \).

Work in second part: \( W_2 = \frac{1}{2} (90 + 40) \times 7 = 65 \times 7 = 455 \, \text{J} \).

Total work: \( W = 270 + 455 = 725 \, \text{J} \).

700 J
725 J
750 J
775 J
2

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