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 $
BETA
