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  4. A block of mass m = 2 kg is placed in equilibrium on a moving plank accelerating with acceleration a = 4 m/s2 If coefficient of friction between plank and block μ=0.2 The frictional force acting on the block is:

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Q. A block of mass $m = 2 \,kg$ is placed in equilibrium on a moving plank accelerating with acceleration $a = 4 \,m/s^{2}$ If coefficient of friction between plank and block $\mu=0.2$ The frictional force acting on the block is:

Laws of Motion

Solution:

Driving force w.r.t. platform = ma
$=2 \times 4=8\,N$
And resisting force $=f_{\text{lim}} = \mu\,N$
$=0.2\times 2 \times 10 =4\,N$
Here resisting force is sufficient to keep the block at rest w.r.t plank hence the block will slide on plank and friction force will be $4\,N$