Work Energy And Power Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A motor lifts a \( 1700 \, \text{kg} \) load at \( 4 \, \text{m/s} \) against \( 4500 \, \text{N} \) friction. What is the power? (Take \( g = 10 \, \text{m/s}^2 \))

Force \( F = mg + F_f = 1700 \times 10 + 4500 = 21500 \, \text{N} \).

Power \( P = F \cdot v = 21500 \times 4 = 86000 \, \text{W} \).

84 kW
86 kW
88 kW
90 kW
2

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

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

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

Scalar product: \( W = (5 \times 2) + (3 \times 4) = 10 + 12 = 22 \, \text{J} \).

18 J
20 J
22 J
24 J
3

A spring (\( k = 300 \, \text{N/m} \)) is compressed by \( 0.3 \, \text{m} \). What is the stored potential energy?

Potential energy \( V = \frac{1}{2} k x^2 = \frac{1}{2} \times 300 \times (0.3)^2 = 150 \times 0.09 = 13.5 \, \text{J} \).

12 J
13.5 J
15 J
16.5 J
2

A \( 0.5 \, \text{g} \) drop falls from \( 600 \, \text{m} \) and hits the ground at \( 20 \, \text{m/s} \). What is the work done by the resistive force? (Take \( g = 10 \, \text{m/s}^2 \))

Work by gravity \( W_g = mgh = 0.0005 \times 10 \times 600 = 3 \, \text{J} \).

Final \( K = \frac{1}{2} \times 0.0005 \times 20^2 = 0.1 \, \text{J} \).

By work-energy theorem, \( K_f = W_g + W_r \Rightarrow 0.1 = 3 + W_r \Rightarrow W_r = -2.9 \, \text{J} \).

-3.2 J
-2.9 J
-2.5 J
-2 J
2

A motor lifts a \( 1800 \, \text{kg} \) load at \( 2.8 \, \text{m/s} \) against \( 3500 \, \text{N} \) friction. What is the power? (Take \( g = 10 \, \text{m/s}^2 \))

Force \( F = mg + F_f = 1800 \times 10 + 3500 = 21500 \, \text{N} \).

Power \( P = F \cdot v = 21500 \times 2.8 = 60200 \, \text{W} \).

58 kW
60 kW
62 kW
64 kW
2

A \( 0.9 \, \text{kg} \) object is dropped from \( 20 \, \text{m} \). What is its kinetic energy just before impact? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( V = mgh = 0.9 \times 10 \times 20 = 180 \, \text{J} \).

By conservation, \( K_f = V_i = 180 \, \text{J} \).

160 J
180 J
200 J
220 J
2

A force \( \mathbf{F} = 4\hat{\mathbf{i}} + 6\hat{\mathbf{j}} \, \text{N} \) acts on a particle moving along \( \mathbf{d} = -3\hat{\mathbf{i}} + 2\hat{\mathbf{j}} \, \text{m} \). What is the work done?

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

\( \mathbf{F} = 4\hat{\mathbf{i}} + 6\hat{\mathbf{j}} \), \( \mathbf{d} = -3\hat{\mathbf{i}} + 2\hat{\mathbf{j}} \).

Scalar product: \( W = (4 \times -3) + (6 \times 2) = -12 + 12 = 0 \, \text{J} \).

-2 J
0 J
2 J
4 J
2

A \( 7 \, \text{kg} \) mass falls from \( 3 \, \text{m} \) onto a spring (\( k = 2000 \, \text{N/m} \)). What is the maximum compression? (Take \( g = 10 \, \text{m/s}^2 \))

Potential energy \( mgh = 7 \times 10 \times 3 = 210 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 210 \Rightarrow 1000 x_m^2 = 210 \Rightarrow x_m = \sqrt{0.21} \approx 0.458 \, \text{m} \).

0.4 m
0.46 m
0.5 m
0.55 m
2

A \( 1000 \, \text{kg} \) car at \( 10 \, \text{m/s} \) hits a spring (\( k = 5 \times 10^3 \, \text{N/m} \)). What is the maximum compression?

Initial \( K = \frac{1}{2} \times 1000 \times 10^2 = 50000 \, \text{J} \).

Spring energy \( \frac{1}{2} k x_m^2 = 50000 \Rightarrow 2500 x_m^2 = 50000 \Rightarrow x_m = \sqrt{20} \approx 4.47 \, \text{m} \).

4 m/s
4.5 m/s
5 m/s
5.5 m/s
2

A \( 1 \, \text{kg} \) ball at \( 10 \, \text{m/s} \) collides elastically with a stationary \( 2 \, \text{kg} \) ball. What is the speed of the \( 2 \, \text{kg} \) ball after collision?

For elastic collision: \( v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i} = \frac{2 \times 1}{1 + 2} \times 10 = \frac{20}{3} \approx 6.67 \, \text{m/s} \).

6 m/s
6.7 m/s
7 m/s
7.5 m/s
2

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