Question

Suppose the block of the previous problem is pusshed down the incline with a force of 4 N. How far will the block move in the first two seconds after starting from rest ? The mass of the block is 4 kg.

Answer 1

From the free body diagram
$4-4a-\mu R+4g\sin {30}^{\circ }=0R-4g\cos {30}^{\circ }=0\Rightarrow R=4g\cos {30}^{\circ }=0$

Putting the values of R in equation (i) we get,
$4-0.11\times 4g\cos {30}^{\circ }+4g\sin {30}^{\circ }=0\Rightarrow 4-4a-0.11\times 4\times 10\times \frac{\sqrt{3}}{2}+4\times 10\times \frac{1}{2}\Rightarrow 4-4a-3.81+20=0\Rightarrow 4-4a-3.18+20=0\Rightarrow a\approx 5m/s^2$
For the block,$u=0,t=2sec,a=5/s^2$
$\therefore Distance,S=ut+\frac{1}{2}a{t}^2=0+\left(\frac{1}{2}\right)5\times 2^2=10m$
Hence the block will move 10 m.