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} $



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