First slide

Second law

Question

A 60 kg mass is pushed with a enough force to start it moving and the same force is continued to act
afterwards. If the coefficient of static friction and sliding friction are 0.5 and 0.4 respectively, then the
acceleration of the body will be [BCECE (Mains) 2012]

Moderate

A

1 $m s^{- 2}$
B

3.9 $m s^{- 2}$
C

4.9 $m s^{- 2}$
D

6 $m s^{- 2}$

Solution

Kinetic friction force $= μ_{k} R = μ_{k} m g = 0.4 \times 60 \times 10 = 240 N$ and the limiting friction force
$= μ_{s} R = μ_{s} m g = 0.5 \times 60 \times 10 = 300 N$
So, the force applied on the body is 300 N and if the body is moving afterwards with the same force, then,
Net accelerating force = Applied force - Kinetic force
$m a = 300 - 240$
$\Rightarrow a = \frac{60}{m} = \frac{60}{60} = 1 {ms}^{- 2}$

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