NEET Laws of Motion — Set 6

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Question 1

A $ 3.2 \, \text{kg} $ block on a $ 45^\circ $ incline ($ \mu_s = 0.45 $) is pulled upward by a horizontal force just sufficient to start motion. What is the force? (Take $ g = 10 \, \text{m/s}^2 $, $ \sin 45^\circ = 0.707 $, $ \cos 45^\circ = 0.707 $)

TopperSquare Expert · Accepted Answer

The block is on the verge of moving upward, so friction acts downward along the incline. Along incline: $ F \cos 45^\circ - mg \sin 45^\circ - f_s = 0 $ (horizontal force component opposes gravity and friction). Normal force: $ N = mg \cos 45^\circ + F \sin 45^\circ $ (horizontal force increases normal force). Calculate: $ mg = 3.2 \times 10 = 32 \, \text{N} $, $ mg \sin 45^\circ = 32 \times 0.707 = 22.624 \, \text{N} $. $ mg \cos 45^\circ = 32 \times 0.707 = 22.624 \, \text{N} $, so $ N = 22.624 + F \times 0.707 $. Friction: $ f_s = \mu_s N = 0.45 (22.624 + 0.707F) $. Substitute: $ F \times 0.707 - 22.624 - 0.45 (22.624 + 0.707F) = 0 $. Expand: $ 0.707F - 22.624 - 10.1808 - 0.31815F = 0 \Rightarrow 0.38885F - 32.8048 = 0 $. Solve: $ 0.38885F = 32.8048 \Rightarrow F \approx \frac{32.8048}{0.38885} \approx 84.34 \, \text{N} $.

Question 2

A $ 900 \, \text{kg} $ car turns on a banked road ($ \theta = 30^\circ $, $ \mu_s = 0.3 $) with radius $ 45 \, \text{m} $. What is the maximum speed without slipping? (Take $ g = 10 \, \text{m/s}^2 $, $ \tan 30^\circ = 0.577 $)

TopperSquare Expert · Accepted Answer

Maximum speed: $ v_{\text{max}} = \sqrt{rg \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}} $. $ \mu_s + \tan\theta = 0.3 + 0.577 = 0.877 $. $ 1 - 0.3 \times 0.577 = 1 - 0.1731 = 0.8269 $. $ v_{\text{max}}^2 = 45 \times 10 \times \frac{0.877}{0.8269} \approx 450 \times 1.0606 \approx 477.27 $. $ v_{\text{max}} \approx \sqrt{477.27} \approx 21.85 \, \text{m/s} $.

Question 3

A $ 0.44 \, \text{kg} $ stone is whirled in a horizontal circle of radius $ 1.6 \, \text{m} $ with a tension of $ 15 \, \text{N} $. What is the speed? (Take $ g = 10 \, \text{m/s}^2 $)

TopperSquare Expert · Accepted Answer

Tension provides centripetal force: $ T = m v^2 / r $. Substitute: $ 15 = 0.44 \times v^2 / 1.6 $. $ 15 = 0.275 v^2 \Rightarrow v^2 = \frac{15}{0.275} \approx 54.55 $. $ v = \sqrt{54.55} \approx 7.39 \, \text{m/s} $.

NEET Physics — Laws of Motion

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NEET Physics: Laws of Motion — MCQ Test 6

Q6. A \( 0.2 tare \text{kg} \) stone is whirled in a circle of radius \( 1.5 \, \text{m} \) with a string that breaks at \( 100 \, \text{N} \). What is the maximum angular speed? (Take \( g = 10 \, \text{m/s}^2 \))

Q7. A \( 1000 \, \text{kg} \) car turns on a banked road (\( \theta = 15^\circ \), \( \mu_s = 0.2 \)) with radius \( 50 \, \text{m} \). What is the maximum speed without slipping? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 15^\circ \approx 0.268 \))

Q8. A \( 1150 \, \text{kg} \) car turns on a banked road (\( \theta = 30^\circ \), \( \mu_s = 0.25 \)) with radius \( 35 \, \text{m} \). What is the maximum speed without slipping? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 30^\circ = 0.577 \))

Q10. A \( 0.29 \, \text{kg} \) ball moving at \( 18 \, \text{m/s} \) strikes a wall perpendicularly and rebounds at \( 14 \, \text{m/s} \). What is the impulse? (Take \( g = 10 \, \text{m/s}^2 \))