SATHEE: Laws Of Motion Question 124

Laws Of Motion Question 124

Question: A block of mass $M = 5 k g$ is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force $F = 40 N$ is applied, the acceleration of the block will be $(g = 10 m / s^{2})$ [MP PMT 2004]

Options:

A) $5.73 m / \sec^{2}$

B) $8.0 m / \sec^{2}$

C) $3.17 m / \sec^{2}$

D) $10.0 m / \sec^{2}$

Show Answer

Answer:

Correct Answer: A

Solution:

Kinetic friction = $μ_{k} R$

$= 0.2 (m g - F \sin 30^{\circ})$

$= 0.2 (5 \times 10 - 40 \times \frac{1}{2})$

$= 0.2 (50 - 20) = 6 N$

Acceleration of the block $= \frac{F \cos 30^{\circ} - Kinetic friction}{,}$

$= \frac{40 \times \frac{\sqrt{3}}{2} - 6}{5} = 5.73 m / s^{2}$

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Laws Of Motion Question 124

Question: A block of mass M=5kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F=40N is applied, the acceleration of the block will be (g=10m/s2) [MP PMT 2004]

Options:

Answer:

Solution:

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Question: A block of mass $M = 5 k g$ is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force $F = 40 N$ is applied, the acceleration of the block will be $(g = 10 m / s^{2})$ [MP PMT 2004]