SATHEE: Laws Of Motion Question 100

Laws Of Motion Question 100

Question: A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and $g = 9.8 m / s^{2}$ , then the acceleration of the block is

Options:

A) $0.26 m / s^{2}$

B) $0.39 m / s^{2}$

C) $0.69 m / s^{2}$

D) $0.88 m / s^{2}$

Show Answer

Answer:

Correct Answer: D

Solution:

Net force = Applied force ? Friction force $m a = 24 - μ m g$

$= 24 - 0.4 \times 5 \times 9.8$

$= 24 - 19.6$

therefore $a = \frac{4.4}{5} = 0.88 m / s^{2}$

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Laws Of Motion Question 100

Question: A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and g=9.8m/s2 , then the acceleration of the block is

Options:

Answer:

Solution:

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Question: A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and $g = 9.8 m / s^{2}$ , then the acceleration of the block is