Work Energy and Power Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A force \( F = 6x^2 \, \text{N} \) acts from \( x = 0 \) to \( x = 1 \, \text{m} \). What is the work done?

Work \( W = \int_0^1 6x^2 \, dx = \left[ 2x^3 \right]_0^1 = 2 \times 1 - 0 = 2 \, \text{J} \).

1.5 J
2 J
2.5 J
3 J
2

A \( 3 \, \text{kg} \) block at \( 8 \, \text{m/s} \) encounters a force \( F = \frac{-1.2}{x} \, \text{N} \) from \( x = 0.5 \, \text{m} \) to \( x = 2.5 \, \text{m} \). What is the final speed?

Initial \( K = \frac{1}{2} \times 3 \times 8^2 = 96 \, \text{J} \).

Work \( W = \int_{0.5}^{2.5} \frac{-1.2}{x} \, dx = -1.2 \ln(2.5/0.5) = -1.2 \ln 5 \approx -1.93 \, \text{J} \).

Final \( K = 96 - 1.93 = 94.07 \, \text{J} \Rightarrow v_f = \sqrt{\frac{2 \times 94.07}{3}} \approx 7.92 \, \text{m/s} \).

7.7 m/s
7.9 m/s
8.1 m/s
8.3 m/s
2

A \( 9 \, \text{kg} \) mass at \( 7 \, \text{m/s} \) collides inelastically with a stationary \( 3 \, \text{kg} \) mass. What is the final speed?

Momentum conservation: \( 9 \times 7 = (9 + 3) v_f \Rightarrow 63 = 12 v_f \Rightarrow v_f = 5.25 \, \text{m/s} \).

5 m/s
5.25 m/s
5.5 m/s
6 m/s
2

A \( 1.5 \, \text{kg} \) ball at \( 15 \, \text{m/s} \) collides elastically with a stationary \( 4.5 \, \text{kg} \) ball. What is the speed of the \( 1.5 \, \text{kg} \) ball after collision?

For elastic collision: \( v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} = \frac{1.5 - 4.5}{1.5 + 4.5} \times 15 = \frac{-3}{6} \times 15 = -7.5 \, \text{m/s} \) (magnitude \( 7.5 \, \text{m/s} \)).

7 m/s
7.5 m/s
8 m/s
8.5 m/s
2

A block of mass \( 1 \, \text{kg} \) moves on a rough surface with initial speed \( 4 \, \text{m/s} \). A retarding force \( F = \frac{-0.8}{x} \, \text{N} \) acts from \( x = 0.2 \, \text{m} \) to \( x = 2 \, \text{m} \). What is the final speed?

Initial \( K = \frac{1}{2} \times 1 \times 4^2 = 8 \, \text{J} \).

Work done by force: \( W = \int_{0.2}^{2} \frac{-0.8}{x} \, dx = -0.8 \ln(x) \Big|_{0.2}^{2} = -0.8 (\ln 2 - \ln 0.2) = -0.8 \ln 10 \approx -1.84 \, \text{J} \).

Final \( K = 8 - 1.84 = 6.16 \, \text{J} \).

\( v_f = \sqrt{\frac{2 \times 6.16}{1}} \approx 3.5 \, \text{m/s} \).

3.2 m/s
3.8 m/s
3.5 m/s
4.0 m/s
3

A block of mass \( 2 \, \text{kg} \) is pushed with a constant force of \( 10 \, \text{N} \) over a distance of \( 5 \, \text{m} \) on a frictionless surface. What is the final kinetic energy of the block if it starts from rest?

Work done \( W = F \cdot d = 10 \times 5 = 50 \, \text{J} \).

By work-energy theorem, \( W = \Delta K = K_f - K_i \).

Since \( K_i = 0 \), \( K_f = 50 \, \text{J} \).

40 J
50 J
60 J
70 J
2

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

Work \( W = mgh = 10 \times 10 \times 5 = 500 \, \text{J} \).

Power \( P = \frac{W}{t} = \frac{500}{2} = 250 \, \text{W} \).

225 W
250 W
275 W
300 W
2

A \( 10 \, \text{kg} \) mass at \( 8 \, \text{m/s} \) collides inelastically with a stationary \( 5 \, \text{kg} \) mass. What is the final speed?

Momentum conservation: \( 10 \times 8 = (10 + 5) v_f \Rightarrow 80 = 15 v_f \Rightarrow v_f = \frac{80}{15} \approx 5.33 \, \text{m/s} \).

5 m/s
5.3 m/s
5.6 m/s
6 m/s
2

A force \( F = 5x \, \text{N} \) acts from \( x = 0 \) to \( x = 3 \, \text{m} \). What is the work done?

Work \( W = \int_0^3 5x \, dx = \left[ \frac{5}{2} x^2 \right]_0^3 = \frac{5}{2} \times 9 = 22.5 \, \text{J} \).

20 J
22.5 J
25 J
27.5 J
2

A spring (\( k = 500 \, \text{N/m} \)) is stretched from \( 0.12 \, \text{m} \) to \( 0.18 \, \text{m} \). What is the work done by the spring force?

Work \( W_s = \frac{1}{2} k (x_i^2 - x_f^2) = \frac{1}{2} \times 500 \times (0.12^2 - 0.18^2) = 250 \times (0.0144 - 0.0324) = -4.5 \, \text{J} \).

-5 J
-4.5 J
-4 J
-3.5 J
2

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