NEET Laws of Motion — Set 17

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Question 1

A $ 4.6 \, \text{kg} $ block on a horizontal surface ($ \mu_k = 0.2 $) is connected to a $ 6.6 \, \text{kg} $ mass over a pulley. A $ 9 \, \text{N} $ force aids the $ 4.6 \, \text{kg} $ block. What is the acceleration? (Take $ g = 10 \, \text{m/s}^2 $)

TopperSquare Expert · Accepted Answer

The $ 6.6 \, \text{kg} $ mass descends, pulling the $ 4.6 \, \text{kg} $ block with an aiding force. For $ 6.6 \, \text{kg} $: $ 6.6g - T = 6.6a \Rightarrow 66 - T = 6.6a $. For $ 4.6 \, \text{kg} $: $ T + 9 - f_k = 4.6a $. Normal: $ N = mg = 4.6 \times 10 = 46 \, \text{N} $. Friction: $ f_k = 0.2 \times 46 = 9.2 \, \text{N} $. Net force: $ T + 9 - 9.2 = 4.6a \Rightarrow T - 0.2 = 4.6a $. Solve: $ 66 - T = 6.6a $, $ T - 0.2 = 4.6a $. Substitute: $ 66 - (4.6a + 0.2) = 6.6a \Rightarrow 66 - 0.2 - 4.6a = 6.6a \Rightarrow 65.8 = 11.2a $. $ a = \frac{65.8}{11.2} \approx 5.88 \, \text{m/s}^2 $.

Question 2

A $ 2.8 \, \text{kg} $ block on a $ 37^\circ $ incline ($ \mu_s = 0.5 $) is pulled upward by a horizontal force just sufficient to start motion. What is the force? (Take $ g = 10 \, \text{m/s}^2 $, $ \sin 37^\circ = 0.6 $, $ \cos 37^\circ = 0.8 $)

TopperSquare Expert · Accepted Answer

The block is about to move upward, so friction acts downward along the incline. Along incline: $ F \cos 37^\circ - mg \sin 37^\circ - f_s = 0 $ (horizontal force component opposes gravity and friction). Normal force: $ N = mg \cos 37^\circ + F \sin 37^\circ $ (horizontal force increases normal force). Calculate: $ mg = 2.8 \times 10 = 28 \, \text{N} $, $ mg \sin 37^\circ = 28 \times 0.6 = 16.8 \, \text{N} $. $ mg \cos 37^\circ = 28 \times 0.8 = 22.4 \, \text{N} $, so $ N = 22.4 + F \times 0.6 $. Friction: $ f_s = \mu_s N = 0.5 (22.4 + 0.6F) $. Substitute: $ F \times 0.8 - 16.8 - 0.5 (22.4 + 0.6F) = 0 $. Expand: $ 0.8F - 16.8 - 11.2 - 0.3F = 0 \Rightarrow 0.5F - 28 = 0 $. Solve: $ 0.5F = 28 \Rightarrow F = \frac{28}{0.5} = 56 \, \text{N} $.

Question 3

A $ 0.5 \, \text{kg} $ stone is whirled in a horizontal circle of radius $ 1.5 \, \text{m} $ at $ 40 \, \text{rev/min} $. What is the tension in the string? (Take $ g = 10 \, \text{m/s}^2 $)

TopperSquare Expert · Accepted Answer

Angular speed: $ \omega = 40 \times \frac{2\pi}{60} = \frac{4\pi}{3} \, \text{rad/s} $. $ \omega^2 = \left(\frac{4\pi}{3}\right)^2 = \frac{16\pi^2}{9} \approx 17.55 $. Centripetal force: $ F = m \omega^2 r = 0.5 \times 17.55 \times 1.5 $. $ F \approx 0.5 \times 17.55 \times 1.5 = 13.1625 \, \text{N} \approx 13.2 \, \text{N} $. Tension equals this force in horizontal circular motion.

NEET Physics — Laws of Motion

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NEET Physics: Laws of Motion — MCQ Test 17

Q6. A truck accelerates uniformly at \( 3 \, \text{m/s}^2 \) from rest. After \( 5 \, \text{s} \), a stone is dropped from it. What is the horizontal velocity of the stone at \( t = 6 \, \text{s} \)? (Neglect air resistance)

Q7. A \( 1400 \, \text{kg} \) truck accelerates at \( 2 \, \text{m/s}^2 \) for \( 5 \, \text{s} \), then a \( 1.2 \, \text{kg} \) stone is dropped. What is the horizontal distance traveled by the stone in \( 1.5 \, \text{s} \)? (Neglect air resistance)

Q8. A \( 4 \, \text{kg} \) block on a \( 53^\circ \) incline (\( \mu_k = 0.15 \)) is pulled upward by a \( 6 \, \text{kg} \) mass over a pulley. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 53^\circ = 0.8 \), \( \cos 53^\circ = 0.6 \))

Q9. A \( 0.24 \, \text{kg} \) ball moving at \( 15 \, \text{m/s} \) at \( 45^\circ \) to a wall rebounds at \( 30^\circ \) with the same speed. What is the impulse magnitude? (Take \( \sin 45^\circ = 0.707 \), \( \cos 45^\circ = 0.707 \), \( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \))

Q10. A \( 10 \, \text{kg} \) mass hangs from a rope with a \( 60 \, \text{N} \) horizontal force at its midpoint. If a \( 5 \, \text{kg} \) mass is added, what is the new angle with the vertical? (Take \( g = 10 \, \text{m/s}^2 \))