Work Energy and Power Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A force starts at \( 75 \, \text{N} \) over \( 3 \, \text{m} \), then decreases linearly to \( 15 \, \text{N} \) over \( 6 \, \text{m} \). What is the total work done?

Work in first part: \( W_1 = 75 \times 3 = 225 \, \text{J} \).

Work in second part: \( W_2 = \frac{1}{2} (75 + 15) \times 6 = 45 \times 6 = 270 \, \text{J} \).

Total work: \( W = 225 + 270 = 495 \, \text{J} \).

475 J
495 J
515 J
535 J
2

A \( 0.4 \, \text{g} \) drop falls from \( 400 \, \text{m} \) and hits the ground at \( 25 \, \text{m/s} \). What is the work done by air resistance? (Take \( g = 10 \, \text{m/s}^2 \))

Work by gravity \( W_g = mgh = 0.0004 \times 10 \times 400 = 1.6 \, \text{J} \).

Final \( K = \frac{1}{2} \times 0.0004 \times 25^2 = 0.125 \, \text{J} \).

\( K_f = W_g + W_r \Rightarrow 0.125 = 1.6 + W_r \Rightarrow W_r = -1.475 \, \text{J} \).

-1.6 J
-1.5 J
-1.4 J
-1.2 J
2

A \( 12 \, \text{kg} \) mass is lifted \( 4 \, \text{m} \) in \( 3 \, \text{s} \) at constant speed. What is the power? (Take \( g = 10 \, \text{m/s}^2 \))

Work \( W = mgh = 12 \times 10 \times 4 = 480 \, \text{J} \).

Power \( P = \frac{W}{t} = \frac{480}{3} = 160 \, \text{W} \).

140 W
160 W
180 W
200 W
2

A \( 800 \, \text{kg} \) car at \( 12 \, \text{m/s} \) hits a spring (\( k = 4 \times 10^3 \, \text{N/m} \)). What is the maximum compression?

Initial \( K = \frac{1}{2} \times 800 \times 12^2 = 57600 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 57600 \Rightarrow 2000 x_m^2 = 57600 \Rightarrow x_m = \sqrt{28.8} \approx 5.37 \, \text{m} \).

5 m
5.4 m
5.8 m
6 m
2

A \( 9 \, \text{kg} \) mass is lifted \( 3 \, \text{m} \) in \( 1.5 \, \text{s} \) at constant speed. What is the power? (Take \( g = 10 \, \text{m/s}^2 \))

Work \( W = mgh = 9 \times 10 \times 3 = 270 \, \text{J} \).

Power \( P = \frac{W}{t} = \frac{270}{1.5} = 180 \, \text{W} \).

160 W
180 W
200 W
220 W
2

A raindrop of mass \( 2 \, \text{g} \) falls from a height of \( 1 \, \text{km} \) and hits the ground with a speed of \( 40 \, \text{m/s} \). What is the work done by the resistive force? (Take \( g = 10 \, \text{m/s}^2 \))

Work by gravity \( W_g = mgh = 0.002 \times 10 \times 1000 = 20 \, \text{J} \).

Final kinetic energy \( K_f = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.002 \times 40^2 = 1.6 \, \text{J} \).

By work-energy theorem, \( \Delta K = W_g + W_r \).

\( 1.6 - 0 = 20 + W_r \Rightarrow W_r = 1.6 - 20 = -18.4 \, \text{J} \).

-20 J
-18.4 J
-16 J
-14 J
2

A \( 4 \, \text{kg} \) block at \( 9 \, \text{m/s} \) encounters a force \( F = \frac{-1.5}{x} \, \text{N} \) from \( x = 0.6 \, \text{m} \) to \( x = 3 \, \text{m} \). What is the final speed?

Initial \( K = \frac{1}{2} \times 4 \times 9^2 = 162 \, \text{J} \).

Work \( W = \int_{0.6}^{3} \frac{-1.5}{x} \, dx = -1.5 \ln(3/0.6) = -1.5 \ln 5 \approx -2.41 \, \text{J} \).

Final \( K = 162 - 2.41 = 159.59 \, \text{J} \Rightarrow v_f = \sqrt{\frac{2 \times 159.59}{4}} \approx 8.93 \, \text{m/s} \).

8.7 m/s
8.9 m/s
9.1 m/s
9.3 m/s
2

A force \( \mathbf{F} = 2\hat{\mathbf{i}} + 5\hat{\mathbf{j}} \, \text{N} \) acts on a particle moving along \( \mathbf{d} = 4\hat{\mathbf{i}} - 3\hat{\mathbf{j}} \, \text{m} \). What is the work done?

Work \( W = \mathbf{F} \cdot \mathbf{d} \).

\( \mathbf{F} = 2\hat{\mathbf{i}} + 5\hat{\mathbf{j}} \), \( \mathbf{d} = 4\hat{\mathbf{i}} - 3\hat{\mathbf{j}} \).

Scalar product: \( W = (2 \times 4) + (5 \times -3) = 8 - 15 = -7 \, \text{J} \).

-9 J
-7 J
-5 J
-3 J
2

A car of mass \( 1500 \, \text{kg} \) moving at \( 20 \, \text{m/s} \) collides with a spring of constant \( 10^4 \, \text{N/m} \). What is the maximum compression?

Initial \( K = \frac{1}{2} \times 1500 \times 20^2 = 300000 \, \text{J} \).

At max compression, \( V = K \Rightarrow \frac{1}{2} k x_m^2 = 300000 \).

\( x_m^2 = \frac{600000}{10^4} = 60 \Rightarrow x_m = \sqrt{60} \approx 7.75 \, \text{m} \).

7 m
7.5 m
7.8 m
8 m
3

A \( 0.7 \, \text{g} \) drop falls from \( 250 \, \text{m} \) and hits the ground at \( 16 \, \text{m/s} \). What is the work done by air resistance? (Take \( g = 10 \, \text{m/s}^2 \))

Work by gravity \( W_g = mgh = 0.0007 \times 10 \times 250 = 1.75 \, \text{J} \).

Final \( K = \frac{1}{2} \times 0.0007 \times 16^2 = 0.0896 \, \text{J} \).

\( K_f = W_g + W_r \Rightarrow 0.0896 = 1.75 + W_r \Rightarrow W_r = -1.6604 \, \text{J} \approx -1.66 \, \text{J} \).

-1.8 J
-1.66 J
-1.5 J
-1.4 J
2

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