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Standard IX

Physics

Frictional Force

A body of 10 ...

Question

A body of 10 kg is acted by a force of 129.4 N if g=9.8 $m / s e c^{2}$ . The acceleration of the block is 10 $m / s^{2}$ . What is the coefficient of kinetic friction

0.03

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0.01

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0.30

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0.25

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Solution

The correct option is C
0.30

Net force on the body = Applied force-Friction
$m a = F - μ_{k} m g \Rightarrow μ_{k} = \frac{F - m a}{m g} = \frac{129.4 - 10 \times 10}{10 \times 9.8} = 0.3$

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A body of 10 kg is acted by a force of 129.4 N if g=9.8 m/sec2. The acceleration of the block is 10 m/s2. What is the coefficient of kinetic friction

The correct option is C 0.30 Net force on the body = Applied force-Friction ma=F−μkmg ⇒ μk=F−mamg=129.4−10×1010×9.8=0.3

A body of 10 kg is acted by a force of 129.4 N if g=9.8 $m / s e c^{2}$ . The acceleration of the block is 10 $m / s^{2}$ . What is the coefficient of kinetic friction

The correct option is C
0.30

Net force on the body = Applied force-Friction
$m a = F - μ_{k} m g \Rightarrow μ_{k} = \frac{F - m a}{m g} = \frac{129.4 - 10 \times 10}{10 \times 9.8} = 0.3$