Laws Of Motion Question 119
Question: A horizontal force of 129.4 N is applied on a 10 kg block which rests on a horizontal surface. If the coefficient of friction is 0.3, the acceleration should be
Options:
A) $ 9.8m/s^{2} $
B) $ 10m/s^{2} $
C) $ 12.6m/s^{2} $
D) $ 19.6m/s^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
From the relation $ F-\mu mg=ma $
$ a=\frac{F-\mu mg}{m}=\frac{129.4-0.3\times 10\times 9.8}{10}=10\ m/s^{2} $
