Laws Of Motion Question 108

Question: A body of 10 kg is acted by a force of 129.4 N if $ g=9.8m/{{\sec }^{2}} $ . The acceleration of the block is$ 10m/s^{2} $ . What is the coefficient of kinetic friction[EAMCET 1994]

Options:

A) 0.03

B) 0.01

C) 0.30

D) 0.25

Show Answer

Answer:

Correct Answer: C

Solution:

Net force on the body = Applied force ? Friction $ ma=F-{{\mu } _{k}}mg $ therefore$ {{\mu } _{k}}=\frac{F-ma}{mg}=\frac{129.4-10\times 10}{10\times 9.8}=0.3 $



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