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/s^{2} $ , then the acceleration of the block is
Options:
A) $ 0.26m/s^{2} $
B) $ 0.39m/s^{2} $
C) $ 0.69m/s^{2} $
D) $ 0.88m/s^{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
Net force = Applied force ? Friction force $ ma=24-\mu \ mg $
$ =24-0.4\times 5\times 9.8 $
$ =24-19.6 $
therefore $ a=\frac{4.4}{5}=0.88\ m/s^{2} $
BETA
