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  4. Work Energy And Power Chapter-Wise Test 2

Work Energy And Power Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 1300 \, \text{kg} \) car at \( 18 \, \text{m/s} \) hits a spring (\( k = 8 \times 10^3 \, \text{N/m} \)). What is the maximum compression?

Initial \( K = \frac{1}{2} \times 1300 \times 18^2 = 210600 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 210600 \Rightarrow 4000 x_m^2 = 210600 \Rightarrow x_m = \sqrt{52.65} \approx 7.26 \, \text{m} \).

7 m
7.3 m
7.5 m
7.8 m
2

A force starts at \( 85 \, \text{N} \) over \( 5 \, \text{m} \), then decreases linearly to \( 25 \, \text{N} \) over \( 5 \, \text{m} \). What is the total work done?

Work in first part: \( W_1 = 85 \times 5 = 425 \, \text{J} \).

Work in second part: \( W_2 = \frac{1}{2} (85 + 25) \times 5 = 55 \times 5 = 275 \, \text{J} \).

Total work: \( W = 425 + 275 = 700 \, \text{J} \).

675 J
700 J
725 J
750 J
2

A \( 0.5 \, \text{g} \) drop falls from \( 350 \, \text{m} \) and hits the ground at \( 17 \, \text{m/s} \). What is the work done by air resistance? (Take \( g = 10 \, \text{m/s}^2 \))

Work by gravity \( W_g = mgh = 0.0005 \times 10 \times 350 = 1.75 \, \text{J} \).

Final \( K = \frac{1}{2} \times 0.0005 \times 17^2 = 0.07225 \, \text{J} \).

\( K_f = W_g + W_r \Rightarrow 0.07225 = 1.75 + W_r \Rightarrow W_r = -1.67775 \, \text{J} \approx -1.68 \, \text{J} \).

-1.8 J
-1.68 J
-1.5 J
-1.4 J
2

A spring (\( k = 400 \, \text{N/m} \)) is stretched by \( 0.05 \, \text{m} \). What is the stored potential energy?

Potential energy \( V = \frac{1}{2} k x^2 = \frac{1}{2} \times 400 \times (0.05)^2 = 200 \times 0.0025 = 0.5 \, \text{J} \).

0.25 J
0.5 J
0.75 J
1 J
2

A spring (\( k = 350 \, \text{N/m} \)) is stretched by \( 0.25 \, \text{m} \). What is the stored potential energy?

Potential energy \( V = \frac{1}{2} k x^2 = \frac{1}{2} \times 350 \times (0.25)^2 = 175 \times 0.0625 = 10.9375 \, \text{J} \approx 10.94 \, \text{J} \).

10 J
10.9 J
11.5 J
12 J
2

A \( 2 \, \text{kg} \) block with initial speed \( 5 \, \text{m/s} \) encounters a force \( F = \frac{-1}{x} \, \text{N} \) from \( x = 0.5 \, \text{m} \) to \( x = 2.5 \, \text{m} \). What is the final speed?

Initial \( K = \frac{1}{2} \times 2 \times 5^2 = 25 \, \text{J} \).

Work \( W = \int_{0.5}^{2.5} \frac{-1}{x} \, dx = -\ln(x) \Big|_{0.5}^{2.5} = -\ln(2.5/0.5) = -\ln 5 \approx -1.61 \, \text{J} \).

Final \( K = 25 - 1.61 = 23.39 \, \text{J} \Rightarrow v_f = \sqrt{\frac{2 \times 23.39}{2}} \approx 4.83 \, \text{m/s} \).

4.5 m/s
4.8 m/s
5.0 m/s
5.2 m/s
2

A motor lifts a \( 1900 \, \text{kg} \) load at \( 3 \, \text{m/s} \) against \( 5000 \, \text{N} \) friction. What is the power? (Take \( g = 10 \, \text{m/s}^2 \))

Force \( F = mg + F_f = 1900 \times 10 + 5000 = 24000 \, \text{N} \).

Power \( P = F \cdot v = 24000 \times 3 = 72000 \, \text{W} \).

70 kW
72 kW
74 kW
76 kW
2

A force \( F = 4x^2 \, \text{N} \) acts from \( x = 0 \) to \( x = 2 \, \text{m} \). What is the work done?

Work \( W = \int_0^2 4x^2 \, dx = \left[ \frac{4}{3} x^3 \right]_0^2 = \frac{4}{3} \times 8 = \frac{32}{3} \approx 10.67 \, \text{J} \).

10 J
10.7 J
11 J
12 J
2

A force starts at \( 70 \, \text{N} \) over \( 4 \, \text{m} \), then decreases linearly to \( 30 \, \text{N} \) over \( 8 \, \text{m} \). What is the total work done?

Work in first part: \( W_1 = 70 \times 4 = 280 \, \text{J} \).

Work in second part: \( W_2 = \frac{1}{2} (70 + 30) \times 8 = 50 \times 8 = 400 \, \text{J} \).

Total work: \( W = 280 + 400 = 680 \, \text{J} \).

650 J
680 J
710 J
740 J
2

A force \( F = 12x \, \text{N} \) acts from \( x = 0 \) to \( x = 3 \, \text{m} \). What is the work done?

Work \( W = \int_0^3 12x \, dx = \left[ 6x^2 \right]_0^3 = 6 \times 9 = 54 \, \text{J} \).

50 J
54 J
58 J
62 J
2

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