Laws of Motion Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A \( 1500 \, \text{kg} \) car on a banked road (\( \theta = 20^\circ \), \( \mu_s = 0.25 \)) turns at radius \( 60 \, \text{m} \) with a tangential force of \( 3000 \, \text{N} \) reducing speed. What is the maximum initial speed? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 20^\circ \approx 0.364 \))

Max speed: \( v_{\text{max}} = \sqrt{rg \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}} \) (radial friction inward).

\( \mu_s + \tan\theta = 0.25 + 0.364 = 0.614 \), \( 1 - 0.25 \times 0.364 = 1 - 0.091 = 0.909 \).

\( v_{\text{max}}^2 = 60 \times 10 \times \frac{0.614}{0.909} \approx 600 \times 0.676 \approx 405.6 \).

\( v_{\text{max}} \approx \sqrt{405.6} \approx 20.14 \, \text{m/s} \).

Tangential force affects deceleration, not max speed limit.

18 m/s
20.1 m/s
22 m/s
24 m/s
2

A \( 0.36 \, \text{kg} \) stone is whirled in a horizontal circle of radius \( 1.7 \, \text{m} \) with a tension of \( 14 \, \text{N} \). What is the speed? (Take \( g = 10 \, \text{m/s}^2 \))

Tension provides centripetal force: \( T = m v^2 / r \).

Substitute: \( 14 = 0.36 \times v^2 / 1.7 \).

\( 14 = 0.21176 v^2 \Rightarrow v^2 = \frac{14}{0.21176} \approx 66.13 \).

\( v = \sqrt{66.13} \approx 8.13 \, \text{m/s} \).

7.5 m/s
8.1 m/s
8.8 m/s
9.5 m/s
2

A \( 6 \, \text{kg} \) block on a \( 53^\circ \) incline (\( \mu_s = 0.6 \)) is pulled upward by a rope at \( 37^\circ \) to the incline with a force just sufficient to initiate motion. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 53^\circ = 0.8 \), \( \cos 53^\circ = 0.6 \), \( \sin 37^\circ = 0.6 \), \( \cos 37^\circ = 0.8 \))

Since the block is about to move upward, the net force along the incline must be zero just before motion starts.

Resolve forces along the incline: \( T \cos 37^\circ \) (upward component of tension) opposes \( mg \sin 53^\circ \) (gravity component down the incline) and \( f_s \) (static friction, also down since motion is impending upward).

Equation: \( T \cos 37^\circ - mg \sin 53^\circ - f_s = 0 \).

Normal force: \( N = mg \cos 53^\circ + T \sin 37^\circ \) (tension’s perpendicular component adds to normal force).

Calculate: \( mg = 6 \times 10 = 60 \, \text{N} \).

\( mg \sin 53^\circ = 60 \times 0.8 = 48 \, \text{N} \), \( mg \cos 53^\circ = 60 \times 0.6 = 36 \, \text{N} \).

So, \( N = 36 + T \times 0.6 \).

Maximum static friction: \( f_s = \mu_s N = 0.6 (36 + 0.6T) \).

Substitute into the incline equation: \( T \times 0.8 - 48 - 0.6 (36 + 0.6T) = 0 \).

Expand: \( 0.8T - 48 - (21.6 + 0.36T) = 0 \).

Simplify: \( 0.8T - 48 - 21.6 - 0.36T = 0 \Rightarrow 0.44T - 69.6 = 0 \).

Solve: \( 0.44T = 69.6 \Rightarrow T = \frac{69.6}{0.44} \approx 158.18 \, \text{N} \).

140 N
158.2 N
170 N
180 N
2

A stone of mass \( 0.3 \, \text{kg} \) is whirled in a horizontal circle of radius \( 2 \, \text{m} \) at \( 20 \, \text{rev/min} \). What is the tension in the string?

Tension provides the centripetal force \( F = m \omega^2 r \).

Convert revolutions to angular speed: \( \omega = 20 \, \text{rev/min} = \frac{20 \times 2\pi}{60} = \frac{2\pi}{3} \, \text{rad/s} \).

Calculate \( \omega^2 = \left(\frac{2\pi}{3}\right)^2 = \frac{4\pi^2}{9} \).

Substitute: \( m = 0.3 \, \text{kg} \), \( r = 2 \, \text{m} \).

So, \( F = 0.3 \times \frac{4\pi^2}{9} \times 2 = 0.3 \times \frac{8\pi^2}{9} \approx 2.63 \, \text{N} \).

1.5 N
2.6 N
3.5 N
4.0 N
2

A \( 0.1 \, \text{kg} \) ball moving at \( 20 \, \text{m/s} \) hits a wall at \( 30^\circ \) and rebounds at the same speed and angle. What is the impulse along the wall? (Take \( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \))

Along wall (y-axis): initial \( p_y = m v \sin 30^\circ = 0.1 \times 20 \times 0.5 = 1 \, \text{kg m/s} \).

Final \( p_y = 1 \, \text{kg m/s} \) (symmetric rebound, no change in y-direction).

Impulse \( \Delta p_y = 1 - 1 = 0 \, \text{N s} \).

Perpendicular (x-axis) changes, but along wall remains zero.

0 N s
1 N s
2 N s
3 N s
1

A \( 6 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.4 \)) is pulled by a \( 8 \, \text{kg} \) mass over a pulley. A \( 25 \, \text{N} \) force at \( 60^\circ \) to the horizontal aids the \( 6 \, \text{kg} \) block, and a \( 15 \, \text{N} \) force opposes the \( 8 \, \text{kg} \) mass. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \))

For \( 8 \, \text{kg} \): \( 8g - T - 15 = 8a \Rightarrow 80 - T - 15 = 8a \Rightarrow 65 - T = 8a \).

For \( 6 \, \text{kg} \): \( T + F \cos 60^\circ - f_k = 6a \).

Normal: \( N = mg - F \sin 60^\circ = 6 \times 10 - 25 \times 0.866 = 60 - 21.65 = 38.35 \, \text{N} \).

Friction: \( f_k = 0.4 \times 38.35 \approx 15.34 \, \text{N} \).

\( F \cos 60^\circ = 25 \times 0.5 = 12.5 \, \text{N} \).

Net force: \( T + 12.5 - 15.34 = 6a \Rightarrow T - 2.84 = 6a \).

Solve: \( 65 - T = 8a \), \( T - 2.84 = 6a \).

Substitute: \( 65 - (6a + 2.84) = 8a \Rightarrow 65 - 2.84 - 6a = 8a \Rightarrow 62.16 = 14a \).

\( a = \frac{62.16}{14} \approx 4.44 \, \text{m/s}^2 \).

\( T - 2.84 = 6 \times 4.44 \Rightarrow T - 2.84 \approx 26.64 \Rightarrow T \approx 29.48 \, \text{N} \).

27 N
29.5 N
32 N
35 N
2

A car of mass \( 800 \, \text{kg} \) takes a circular turn of radius \( 20 \, \text{m} \) at \( 10 \, \text{m/s} \) on a level road. What is the frictional force providing the centripetal force?

Centripetal force \( F_c = \frac{mv^2}{r} \).

Substitute: \( m = 800 \, \text{kg} \), \( v = 10 \, \text{m/s} \), \( r = 20 \, \text{m} \).

\( F_c = \frac{800 \times (10)^2}{20} \).

\( F_c = \frac{800 \times 100}{20} = 4000 \, \text{N} \).

On a level road, friction provides this force, so \( f = 4000 \, \text{N} \).

2000 N
3000 N
4000 N
5000 N
3

A \( 5 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.3 \)) is connected to a \( 3 \, \text{kg} \) mass over a pulley. A force of \( 20 \, \text{N} \) at \( 60^\circ \) to the horizontal pulls the \( 5 \, \text{kg} \) block. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \))

For \( 3 \, \text{kg} \): \( 3g - T = 3a \Rightarrow 30 - T = 3a \).

For \( 5 \, \text{kg} \): \( T + F \cos 60^\circ - f_k = 5a \), \( N = mg - F \sin 60^\circ \).

\( N = 5 \times 10 - 20 \times 0.866 = 50 - 17.32 = 32.68 \, \text{N} \).

\( f_k = 0.3 \times 32.68 \approx 9.8 \, \text{N} \), \( F \cos 60^\circ = 20 \times 0.5 = 10 \, \text{N} \).

\( T + 10 - 9.8 = 5a \Rightarrow T + 0.2 = 5a \).

Solve: \( 30 - T = 3a \), \( T + 0.2 = 5a \Rightarrow 30 - (5a - 0.2) = 3a \Rightarrow 30.2 = 8a \Rightarrow a \approx 3.78 \, \text{m/s}^2 \).

3 m/s²
3.5 m/s²
3.78 m/s²
4 m/s²
3

A \( 55 \, \text{kg} \) man in a lift accelerating upward at \( 1 \, \text{m/s}^2 \) stands on a scale. What is the reading? (Take \( g = 10 \, \text{m/s}^2 \))

Scale reads normal force.

Net force: \( N - mg = ma \).

\( N - 55 \times 10 = 55 \times 1 \).

\( N - 550 = 55 \Rightarrow N = 605 \, \text{N} \).

575 N
605 N
625 N
650 N
2

A \( 0.5 \, \text{kg} \) stone is whirled in a vertical circle of radius \( 2 \, \text{m} \) at \( 30 \, \text{rev/min} \). What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

\( \omega = 30 \times \frac{2\pi}{60} = \pi \, \text{rad/s} \), \( \omega^2 = \pi^2 \approx 9.87 \).

At top: \( T + mg = m \omega^2 r \).

Centripetal force: \( m \omega^2 r = 0.5 \times 9.87 \times 2 \approx 9.87 \, \text{N} \).

\( mg = 0.5 \times 10 = 5 \, \text{N} \).

\( T + 5 = 9.87 \Rightarrow T = 9.87 - 5 \approx 4.87 \, \text{N} \).

4 N
4.9 N
5.5 N
6 N
2

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0