Work Energy and Power Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

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

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

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

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

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

Fraction transferred \( f_2 = \frac{4 m_1 m_2}{(m_1 + m_2)^2} = \frac{4 \times 1 \times 12}{(1 + 12)^2} = \frac{48}{169} \approx 0.284 \).

0.25
0.28
0.32
0.35
2

A \( 5 \, \text{kg} \) ball at \( 24 \, \text{m/s} \) collides elastically with a stationary \( 15 \, \text{kg} \) ball. What is the speed of the \( 5 \, \text{kg} \) ball after collision?

For elastic collision: \( v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} = \frac{5 - 15}{5 + 15} \times 24 = \frac{-10}{20} \times 24 = -12 \, \text{m/s} \) (magnitude \( 12 \, \text{m/s} \)).

10 m/s
12 m/s
14 m/s
16 m/s
2

A \( 8 \, \text{kg} \) mass at \( 12 \, \text{m/s} \) collides elastically with an identical stationary mass. What is the speed of the second mass after collision?

For equal masses in elastic collision, \( v_{2f} = v_{1i} = 12 \, \text{m/s} \).

10 m/s
11 m/s
12 m/s
13 m/s
3

A \( 5 \, \text{kg} \) mass falls from \( 5 \, \text{m} \) onto a spring (\( k = 1000 \, \text{N/m} \)). What is the maximum compression? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( mgh = 5 \times 10 \times 5 = 250 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 250 \Rightarrow 500 x_m^2 = 250 \Rightarrow x_m = \sqrt{0.5} \approx 0.707 \, \text{m} \).

0.6 m
0.7 m
0.8 m
0.9 m
2

A \( 0.6 \, \text{g} \) drop falls from \( 300 \, \text{m} \) and hits the ground at \( 18 \, \text{m/s} \). What is the work done by air resistance? (Take \( g = 10 \, \text{m/s}^2 \))

Work by gravity \( W_g = mgh = 0.0006 \times 10 \times 300 = 1.8 \, \text{J} \).

Final \( K = \frac{1}{2} \times 0.0006 \times 18^2 = 0.0972 \, \text{J} \).

\( K_f = W_g + W_r \Rightarrow 0.0972 = 1.8 + W_r \Rightarrow W_r = -1.7028 \, \text{J} \approx -1.7 \, \text{J} \).

-1.8 J
-1.7 J
-1.6 J
-1.5 J
2

A \( 5 \, \text{kg} \) block slides down a frictionless incline from \( 6 \, \text{m} \) height. What is its speed at the bottom? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( mgh = 5 \times 10 \times 6 = 300 \, \text{J} \).

Kinetic energy \( \frac{1}{2} m v^2 = 300 \Rightarrow v^2 = 120 \Rightarrow v = \sqrt{120} \approx 10.95 \, \text{m/s} \).

10 m/s
11 m/s
12 m/s
13 m/s
2

A \( 6 \, \text{kg} \) mass at \( 10 \, \text{m/s} \) collides elastically with an identical stationary mass. What is the speed of the second mass after collision?

For equal masses in elastic collision, \( v_{2f} = v_{1i} = 10 \, \text{m/s} \).

8 m/s
9 m/s
10 m/s
11 m/s
3

A \( 0.6 \, \text{kg} \) pendulum bob completes a vertical circle of radius \( 1.2 \, \text{m} \). What is the speed at the top? (Take \( g = 10 \, \text{m/s}^2 \))

At top, minimum speed \( v_C = \sqrt{gL} = \sqrt{10 \times 1.2} = \sqrt{12} \approx 3.46 \, \text{m/s} \).

3 m/s
3.5 m/s
4 m/s
4.5 m/s
2

A \( 5 \, \text{kg} \) mass is lifted \( 10 \, \text{m} \) at constant speed by a motor. What is the power if it takes \( 4 \, \text{s} \)? (Take \( g = 10 \, \text{m/s}^2 \))

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

Power \( P = \frac{W}{t} = \frac{500}{4} = 125 \, \text{W} \).

100 W
125 W
150 W
175 W
2

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