NEET Laws of Motion — Set 7

Question 1

A $ 4 \, \text{kg} $ block on a horizontal surface ($ \mu_k = 0.25 $) is connected to a $ 2 \, \text{kg} $ mass over a pulley. A $ 10 \, \text{N} $ horizontal force opposes the motion. What is the acceleration? (Take $ g = 10 \, \text{m/s}^2 $)

Accepted Answer

For $ 2 \, \text{kg} $: $ 2g - T = 2a \Rightarrow 20 - T = 2a $. For $ 4 \, \text{kg} $: $ T - f_k - 10 = 4a $, $ f_k = 0.25 \times 4 \times 10 = 10 \, \text{N} $. $ T - 10 - 10 = 4a \Rightarrow T - 20 = 4a $. Solve: $ 20 - T = 2a $, $ T - 20 = 4a $. Substitute $ T = 20 - 2a $ into $ 20 - 2a - 20 = 4a \Rightarrow -2a = 4a - 20 $. $ 6a = 20 \Rightarrow a = \frac{20}{6} \approx 3.33 \, \text{m/s}^2 $.

Question 2

A $ 2.5 \, \text{kg} $ block on a horizontal surface ($ \mu_k = 0.3 $) is connected to a $ 4.5 \, \text{kg} $ mass over a pulley. A $ 10 \, \text{N} $ force opposes the $ 2.5 \, \text{kg} $ block. What is the acceleration? (Take $ g = 10 \, \text{m/s}^2 $)

Accepted Answer

The $ 4.5 \, \text{kg} $ mass descends, pulling the $ 2.5 \, \text{kg} $ block against friction and the opposing force. For $ 4.5 \, \text{kg} $: $ 4.5g - T = 4.5a \Rightarrow 45 - T = 4.5a $. For $ 2.5 \, \text{kg} $: $ T - f_k - 10 = 2.5a $. Normal: $ N = mg = 2.5 \times 10 = 25 \, \text{N} $. Friction: $ f_k = 0.3 \times 25 = 7.5 \, \text{N} $. Net force: $ T - 7.5 - 10 = 2.5a \Rightarrow T - 17.5 = 2.5a $. Solve: $ 45 - T = 4.5a $, $ T - 17.5 = 2.5a $. Substitute: $ 45 - (2.5a + 17.5) = 4.5a \Rightarrow 45 - 17.5 - 2.5a = 4.5a \Rightarrow 27.5 = 7a $. $ a = \frac{27.5}{7} \approx 3.93 \, \text{m/s}^2 $.

Question 3

A $ 0.15 \, \text{kg} $ ball moving at $ 30 \, \text{m/s} $ at $ 60^\circ $ to a wall rebounds at the same speed and angle. What is the impulse perpendicular to the wall? (Take $ \sin 60^\circ = 0.866 $, $ \cos 60^\circ = 0.5 $)

Accepted Answer

Perpendicular (x-axis): inicial $ p_x = m v \cos 60^\circ = 0.15 \times 30 \times 0.5 = 2.25 \, \text{kg m/s} $. Final $ p_x = -2.25 \, \text{kg m/s} $ (rebounds symmetrically). Impulse $ \Delta p_x = -2.25 - 2.25 = -4.5 \, \text{N s} $. Magnitude = $ 4.5 \, \text{N s} $.

NEET Physics: Laws of Motion — Practice Set 7

Q1. A \( 4 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.25 \)) is connected to a \( 2 \, \text{kg} \) mass over a pulley. A \( 10 \, \text{N} \) horizontal force opposes the motion. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \))

Q2. A \( 2.5 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.3 \)) is connected to a \( 4.5 \, \text{kg} \) mass over a pulley. A \( 10 \, \text{N} \) force opposes the \( 2.5 \, \text{kg} \) block. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \))

Q3. A \( 0.15 \, \text{kg} \) ball moving at \( 30 \, \text{m/s} \) at \( 60^\circ \) to a wall rebounds at the same speed and angle. What is the impulse perpendicular to the wall? (Take \( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \))

Q4. A \( 3 \, \text{kg} \) block on a \( 53^\circ \) incline (\( \mu_s = 0.5 \)) is pulled upward by a horizontal force just sufficient to start motion. What is the force? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 53^\circ = 0.8 \), \( \cos 53^\circ = 0.6 \))

Q5. A \( 3 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.2 \)) is connected to a \( 4 \, \text{kg} \) mass over a pulley. A \( 15 \, \text{N} \) horizontal force pulls the \( 3 \, \text{kg} \) block. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \))

Q6. A \( 0.5 \, \text{kg} \) stone in a vertical circle of radius \( 1.5 \, \text{m} \) has a tension of \( 25 \, \text{N} \) at the bottom. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

Q8. A \( 0.27 \, \text{kg} \) ball moving at \( 24 \, \text{m/s} \) at \( 53^\circ \) to a wall rebounds at \( 37^\circ \) with the same speed. What is the impulse magnitude? (Take \( \sin 53^\circ = 0.8 \), \( \cos 53^\circ = 0.6 \), \( \sin 37^\circ = 0.6 \), \( \cos 37^\circ = 0.8 \))

Q9. A \( 3.4 \, \text{kg} \) block on a \( 60^\circ \) incline (\( \mu_s = 0.4 \)) is pulled upward by a horizontal force just sufficient to start motion. What is the force? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \))

Q10. A \( 60 \, \text{kg} \) man in a lift accelerating downward at \( 2 \, \text{m/s}^2 \) stands on a scale. What is the reading? (Take \( g = 10 \, \text{m/s}^2 \))