Work Energy and Power Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 2.5 \, \text{kg} \) block at \( 7 \, \text{m/s} \) encounters a force \( F = \frac{-0.9}{x} \, \text{N} \) from \( x = 0.4 \, \text{m} \) to \( x = 2 \, \text{m} \). What is the final speed?

Initial \( K = \frac{1}{2} \times 2.5 \times 7^2 = 61.25 \, \text{J} \).

Work \( W = \int_{0.4}^{2} \frac{-0.9}{x} \, dx = -0.9 \ln(2/0.4) = -0.9 \ln 5 \approx -1.45 \, \text{J} \).

Final \( K = 61.25 - 1.45 = 59.8 \, \text{J} \Rightarrow v_f = \sqrt{\frac{2 \times 59.8}{2.5}} \approx 6.92 \, \text{m/s} \).

6.7 m/s
6.9 m/s
7.1 m/s
7.3 m/s
2

A \( 11 \, \text{kg} \) mass at \( 6 \, \text{m/s} \) collides inelastically with a stationary \( 4 \, \text{kg} \) mass. What is the final speed?

Momentum conservation: \( 11 \times 6 = (11 + 4) v_f \Rightarrow 66 = 15 v_f \Rightarrow v_f = 4.4 \, \text{m/s} \).

4 m/s
4.4 m/s
4.8 m/s
5 m/s
2

A \( 5 \, \text{kg} \) block slides down a frictionless incline from \( 9 \, \text{m} \) height. What is its speed at the bottom? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( mgh = 5 \times 10 \times 9 = 450 \, \text{J} \).

Kinetic energy \( \frac{1}{2} m v^2 = 450 \Rightarrow v^2 = 180 \Rightarrow v = \sqrt{180} \approx 13.42 \, \text{m/s} \).

13 m/s
13.4 m/s
14 m/s
14.5 m/s
2

A pendulum bob of mass \( 2 \, \text{kg} \) is given a horizontal velocity \( v_0 \) at the bottom to just complete a vertical circle of radius \( 1.5 \, \text{m} \). What is \( v_0 \)? (Take \( g = 10 \, \text{m/s}^2 \))

At top, \( v = \sqrt{gL} \), energy at bottom = energy at top.

\( \frac{1}{2} m v_0^2 = \frac{1}{2} m (gL) + 2mgL \).

\( v_0^2 = 5gL = 5 \times 10 \times 1.5 = 75 \Rightarrow v_0 = \sqrt{75} \approx 8.66 \, \text{m/s} \).

7.5 m/s
8 m/s
8.7 m/s
9 m/s
3

A \( 9 \, \text{kg} \) mass at \( 15 \, \text{m/s} \) collides elastically with an identical stationary mass. What is the speed of the first mass after collision?

For equal masses in elastic collision, \( v_{1f} = 0 \, \text{m/s} \) (first mass stops).

0 m/s
5 m/s
10 m/s
12 m/s
1

A neutron (\( 1 \, \text{u} \)) at \( 4 \times 10^6 \, \text{m/s} \) collides elastically with a carbon (\( 12 \, \text{u} \)). What fraction of its kinetic energy is retained?

Fraction retained \( f_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right)^2 = \left( \frac{1 - 12}{1 + 12} \right)^2 = \left( \frac{-11}{13} \right)^2 = \frac{121}{169} \approx 0.716 \).

0.68
0.72
0.75
0.78
2

A \( 1.5 \, \text{kg} \) block slides down a frictionless incline from \( 12 \, \text{m} \) height. What is its speed at the bottom? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( mgh = 1.5 \times 10 \times 12 = 180 \, \text{J} \).

Kinetic energy \( \frac{1}{2} m v^2 = 180 \Rightarrow v^2 = 240 \Rightarrow v = \sqrt{240} \approx 15.49 \, \text{m/s} \).

14 m/s
15 m/s
15.5 m/s
16 m/s
3

A \( 15 \, \text{g} \) bullet at \( 800 \, \text{m/s} \) emerges from a target with \( 25\% \) of its initial kinetic energy. What is its emergent speed?

Initial \( K = \frac{1}{2} \times 0.015 \times 800^2 = 4800 \, \text{J} \).

Final \( K = 0.25 \times 4800 = 1200 \, \text{J} \).

\( \frac{1}{2} \times 0.015 \times v_f^2 = 1200 \Rightarrow v_f^2 = 160000 \Rightarrow v_f = 400 \, \text{m/s} \).

380 m/s
400 m/s
420 m/s
440 m/s
2

A \( 2.5 \, \text{kg} \) mass falls from \( 8 \, \text{m} \) onto a spring (\( k = 1000 \, \text{N/m} \)). What is the maximum compression? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( mgh = 2.5 \times 10 \times 8 = 200 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 200 \Rightarrow 500 x_m^2 = 200 \Rightarrow x_m = \sqrt{0.4} \approx 0.63 \, \text{m} \).

0.5 m
0.6 m
0.63 m
0.7 m
3

A \( 8 \, \text{kg} \) block at \( 14 \, \text{m/s} \) encounters a force \( F = \frac{-3}{x} \, \text{N} \) from \( x = 1.2 \, \text{m} \) to \( x = 6 \, \text{m} \). What is the final speed?

Initial \( K = \frac{1}{2} \times 8 \times 14^2 = 784 \, \text{J} \).

Work \( W = \int_{1.2}^{6} \frac{-3}{x} \, dx = -3 \ln(6/1.2) = -3 \ln 5 \approx -4.83 \, \text{J} \).

Final \( K = 784 - 4.83 = 779.17 \, \text{J} \Rightarrow v_f = \sqrt{\frac{2 \times 779.17}{8}} \approx 13.94 \, \text{m/s} \).

13.7 m/s
13.9 m/s
14.1 m/s
14.3 m/s
2

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