Work Energy and Power Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 4 \, \text{kg} \) mass at \( 15 \, \text{m/s} \) collides elastically with an identical stationary mass. What is the speed of the second mass after collision?

For equal masses in elastic collision, \( v_{2f} = v_{1i} = 15 \, \text{m/s} \).

12 m/s
13.5 m/s
15 m/s
16.5 m/s
3

A \( 3 \, \text{kg} \) mass moving at \( 12 \, \text{m/s} \) 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
4 m/s
6 m/s
8 m/s
1

A neutron (\( 1 \, \text{u} \)) at \( 6 \times 10^6 \, \text{m/s} \) collides elastically with a deuterium (\( 2 \, \text{u} \)). What fraction of its kinetic energy is transferred?

Fraction transferred \( f_2 = \frac{4 m_1 m_2}{(m_1 + m_2)^2} = \frac{4 \times 1 \times 2}{(1 + 2)^2} = \frac{8}{9} \approx 0.889 \).

0.85
0.89
0.92
0.95
2

A spring (\( k = 600 \, \text{N/m} \)) is stretched from \( 0.15 \, \text{m} \) to \( 0.25 \, \text{m} \). What is the work done by the spring force?

Work \( W_s = \frac{1}{2} k (x_i^2 - x_f^2) = \frac{1}{2} \times 600 \times (0.15^2 - 0.25^2) = 300 \times (0.0225 - 0.0625) = -12 \, \text{J} \).

-15 J
-12 J
-10 J
-8 J
2

A motor lifts a \( 2000 \, \text{kg} \) elevator at \( 3 \, \text{m/s} \) against a frictional force of \( 5000 \, \text{N} \). What is the power delivered? (Take \( g = 10 \, \text{m/s}^2 \))

Total force \( F = mg + F_f = 2000 \times 10 + 5000 = 25000 \, \text{N} \).

Power \( P = F \cdot v = 25000 \times 3 = 75000 \, \text{W} \).

70 kW
75 kW
80 kW
85 kW
2

A \( 4 \, \text{kg} \) mass falls from \( 6 \, \text{m} \) onto a spring (\( k = 800 \, \text{N/m} \)). What is the maximum compression? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( mgh = 4 \times 10 \times 6 = 240 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 240 \Rightarrow 400 x_m^2 = 240 \Rightarrow x_m = \sqrt{0.6} \approx 0.775 \, \text{m} \).

0.7 m
0.78 m
0.85 m
0.9 m
2

A \( 6 \, \text{kg} \) 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 \( W = mgh = 6 \times 10 \times 5 = 300 \, \text{J} \).

Power \( P = \frac{W}{t} = \frac{300}{2} = 150 \, \text{W} \).

130 W
150 W
170 W
190 W
2

A \( 1400 \, \text{kg} \) car at \( 22 \, \text{m/s} \) hits a spring (\( k = 1.2 \times 10^4 \, \text{N/m} \)). What is the maximum compression?

Initial \( K = \frac{1}{2} \times 1400 \times 22^2 = 338800 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 338800 \Rightarrow 6000 x_m^2 = 338800 \Rightarrow x_m = \sqrt{56.47} \approx 7.51 \, \text{m} \).

7.3 m
7.5 m
7.7 m
8 m
2

A \( 35 \, \text{g} \) bullet at \( 1200 \, \text{m/s} \) emerges from a block with \( 80\% \) of its initial kinetic energy. What is its emergent speed?

Initial \( K = \frac{1}{2} \times 0.035 \times 1200^2 = 25200 \, \text{J} \).

Final \( K = 0.8 \times 25200 = 20160 \, \text{J} \).

\( \frac{1}{2} \times 0.035 \times v_f^2 = 20160 \Rightarrow v_f^2 = 1152000 \Rightarrow v_f = \sqrt{1152000} \approx 1073.3 \, \text{m/s} \).

1050 m/s
1073 m/s
1100 m/s
1120 m/s
2

A motor lifts a \( 2500 \, \text{kg} \) load at \( 2.5 \, \text{m/s} \) against friction of \( 3000 \, \text{N} \). What is the power output? (Take \( g = 10 \, \text{m/s}^2 \))

Force \( F = mg + F_f = 2500 \times 10 + 3000 = 28000 \, \text{N} \).

Power \( P = F \cdot v = 28000 \times 2.5 = 70000 \, \text{W} \).

65 kW
70 kW
75 kW
80 kW
2

A \( 12 \, \text{kg} \) mass at \( 9 \, \text{m/s} \) collides inelastically with a stationary \( 6 \, \text{kg} \) mass. What is the final speed?

Momentum conservation: \( 12 \times 9 = (12 + 6) v_f \Rightarrow 108 = 18 v_f \Rightarrow v_f = 6 \, \text{m/s} \).

5.5 m/s
6 m/s
6.5 m/s
7 m/s
2

A \( 8 \, \text{kg} \) mass at \( 5 \, \text{m/s} \) collides inelastically with a stationary \( 4 \, \text{kg} \) mass. What is the final speed?

Momentum conservation: \( 8 \times 5 = (8 + 4) v_f \Rightarrow 40 = 12 v_f \Rightarrow v_f = \frac{40}{12} \approx 3.33 \, \text{m/s} \).

3 m/s
3.3 m/s
3.6 m/s
4 m/s
2

A neutron (\( 1 \, \text{u} \)) moving at \( 10^6 \, \text{m/s} \) collides elastically with a deuterium (\( 2 \, \text{u} \)). What fraction of its kinetic energy is lost?

Fraction lost \( f_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right)^2 = \left( \frac{1 - 2}{1 + 2} \right)^2 = \left( \frac{-1}{3} \right)^2 = \frac{1}{9} \).

1/12
1/9
1/6
1/3
2

A force \( \mathbf{F} = 6\hat{\mathbf{i}} + 4\hat{\mathbf{j}} \, \text{N} \) acts on a particle moving along \( \mathbf{d} = -2\hat{\mathbf{i}} + 5\hat{\mathbf{j}} \, \text{m} \). What is the work done?

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

\( \mathbf{F} = 6\hat{\mathbf{i}} + 4\hat{\mathbf{j}} \), \( \mathbf{d} = -2\hat{\mathbf{i}} + 5\hat{\mathbf{j}} \).

Scalar product: \( W = (6 \times -2) + (4 \times 5) = -12 + 20 = 8 \, \text{J} \).

6 J
8 J
10 J
12 J
2

A \( 4 \, \text{kg} \) ball at \( 20 \, \text{m/s} \) collides elastically with a stationary \( 12 \, \text{kg} \) ball. What is the speed of the \( 12 \, \text{kg} \) ball after collision?

For elastic collision: \( v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i} = \frac{2 \times 4}{4 + 12} \times 20 = \frac{8}{16} \times 20 = 10 \, \text{m/s} \).

9 m/s
10 m/s
11 m/s
12 m/s
2

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