NEET Laws of Motion — Set 8

Q: Is this practice set completely free?

Yes, all our chapter-wise practice sets are 100% free. You can practice as many times as you want without any hidden charges.

Q: Are these questions relevant for the latest NEET syllabus?

Absolutely. Our content team regularly updates the question bank to strictly align with the latest NTA NEET syllabus and NCERT textbooks.

Q: Can I view the solutions after the test?

Yes, once you submit the test, you will get an instant scorecard along with the correct answers and detailed explanations for every question.

Question 1

A $ 12 \, \text{kg} $ block on a horizontal surface ($ \mu_k = 0.3 $) is pulled by a $ 10 \, \text{kg} $ mass over a pulley. A $ 35 \, \text{N} $ force at $ 53^\circ $ opposes the $ 12 \, \text{kg} $ block, and a $ 20 \, \text{N} $ force aids the $ 10 \, \text{kg} $ mass. What is the acceleration? (Take $ g = 10 \, \text{m/s}^2 $, $ \sin 53^\circ = 0.8 $, $ \cos 53^\circ = 0.6 $)

TopperSquare Expert · Accepted Answer

For $ 10 \, \text{kg} $: $ 10g + 20 - T = 10a \Rightarrow 100 + 20 - T = 10a \Rightarrow 120 - T = 10a $. For $ 12 \, \text{kg} $: $ T - f_k - F \cos 53^\circ = 12a $. Normal: $ N = mg + F \sin 53^\circ = 12 \times 10 + 35 \times 0.8 = 120 + 28 = 148 \, \text{N} $. Friction: $ f_k = 0.3 \times 148 = 44.4 \, \text{N} $. $ F \cos 53^\circ = 35 \times 0.6 = 21 \, \text{N} $. Net force: $ T - 44.4 - 21 = 12a \Rightarrow T - 65.4 = 12a $. Solve: $ 120 - T = 10a $, $ T - 65.4 = 12a $. Substitute: $ 120 - (12a + 65.4) = 10a \Rightarrow 120 - 65.4 - 12a = 10a \Rightarrow 54.6 = 22a $. $ a = \frac{54.6}{22} \approx 2.48 \, \text{m/s}^2 $.

Question 2

A $ 5 \, \text{kg} $ block on a $ 37^\circ $ incline ($ \mu_s = 0.4 $) is pulled upward by a force at $ 30^\circ $ to the horizontal 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 $, $ \sin 30^\circ = 0.5 $, $ \cos 30^\circ = 0.866 $)

TopperSquare Expert · Accepted Answer

The block is on the verge of moving upward, so the net force along the incline is zero, and friction acts downward. Along incline: $ F \cos (37^\circ - 30^\circ) - mg \sin 37^\circ - f_s = 0 $, where $ \theta = 7^\circ $. Approximate: $ \cos 7^\circ \approx 0.992 $, $ \sin 7^\circ \approx 0.122 $ (for simplicity, use exact if needed). Normal force: $ N = mg \cos 37^\circ - F \sin (37^\circ - 30^\circ) $. $ mg \sin 37^\circ = 5 \times 10 \times 0.6 = 30 \, \text{N} $, $ mg \cos 37^\circ = 5 \times 10 \times 0.8 = 40 \, \text{N} $. $ N = 40 - F \times 0.122 $, $ f_s = 0.4 (40 - 0.122F) $. Substitute: $ F \times 0.992 - 30 - 0.4 (40 - 0.122F) = 0 $. $ 0.992F - 30 - 16 + 0.0488F = 0 \Rightarrow 1.0408F - 46 = 0 $. $ 1.0408F = 46 \Rightarrow F \approx 44.2 \, \text{N} $.

Question 3

A $ 8 \, \text{kg} $ block on a horizontal surface ($ \mu_k = 0.3 $) is pulled by a $ 5 \, \text{kg} $ mass over a pulley. What is the tension in the string? (Take $ g = 10 \, \text{m/s}^2 $)

TopperSquare Expert · Accepted Answer

For $ 5 \, \text{kg} $: $ 5g - T = 5a \Rightarrow 50 - T = 5a $. For $ 8 \, \text{kg} $: $ T - f_k = 8a $. Normal: $ N = mg = 8 \times 10 = 80 \, \text{N} $. Friction: $ f_k = 0.3 \times 80 = 24 \, \text{N} $. Net force: $ T - 24 = 8a $. Solve: $ 50 - T = 5a $, $ T - 24 = 8a $. Substitute: $ 50 - (8a + 24) = 5a \Rightarrow 50 - 24 - 8a = 5a \Rightarrow 26 = 13a $. $ a = 2 \, \text{m/s}^2 $. $ T - 24 = 8 \times 2 \Rightarrow T - 24 = 16 \Rightarrow T = 40 \, \text{N} $.

NEET Physics — Laws of Motion

Accelerate Your NEET Preparation

Question Bank

Practice Tests

NEET PYQs

AI Analytics

About this NEET MCQ Test

Preview of Questions in this Test

Frequently Asked Questions

Is this practice set completely free?

Are these questions relevant for the latest NEET syllabus?

Can I view the solutions after the test?

About TopperSquare

More Physics Chapters for NEET

Explore More NEET Subjects

Everything you need to crack NEET

NEET Physics: Laws of Motion — MCQ Test 8

Q6. A \( 2 \, \text{kg} \) block on a \( 37^\circ \) incline (\( \mu_k = 0.3 \)) accelerates down at \( 2 \, \text{m/s}^2 \). What external force acts parallel to the incline? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 37^\circ = 0.6 \), \( \cos 37^\circ = 0.8 \))

Q7. A \( 4 \, \text{kg} \) mass on a surface (\( \mu_k = 0.2 \)) is pulled by a \( 6 \, \text{kg} \) mass over a pulley. If a \( 5 \, \text{N} \) force opposes the \( 4 \, \text{kg} \) mass, what is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

Q9. A constant retarding force of \( 20 \, \text{N} \) stops a body of mass \( 4 \, \text{kg} \) moving at \( 10 \, \text{m/s} \). How long does it take to stop?

Q10. A \( 0.4 \, \text{kg} \) ball at \( 50 \, \text{m/s} \) hits a wall at \( 45^\circ \) and rebounds at \( 60^\circ \) with speed \( 40 \, \text{m/s} \). What is the impulse along the wall? (Take \( \sin 45^\circ = \cos 45^\circ = 0.707 \), \( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \))