Question

The object of mass M=10.00 kg starts out at the bottom of the inclined plane shown below. It is given an initial speed of 18.00 m/s up the plane. Assuming that the coefficient of kinetic friction between the block and the plane is 0.350, how far up the incline will the block slide before coming to a stop (use any appropriate method)?

Answer 1

I mgsina f 1 mg caso are m I and masine are in Same direction.

frictional force (f) is given as ,

$f=\mu mgcos\theta$, it will be opposite to the upward motion of the block.

Block is projected upwards with initial velocity "u" , hence experiences downward acceleration "a".In this case.

$mgsin\theta +\mu mgcos\theta = ma$

$\Rightarrow a=gsin\theta +\mu gcos\theta$

here , $\theta =30^{\circ}$ and $\mu =0.350$

$\Rightarrow a=9.8\times sin30^{\circ} +0.350\times 9.8\times cos30^{\circ}$

$\Rightarrow a=7.87ms^{-2}$

$\Rightarrow a=-7.87ms^{-2}$, negative sign indicates downward acceleration.

let "S" be the distance moved up on the inclined plane before the block comes to rest.

Now using equation,

$v^{2}-u^{2}=2\times a\times s$

here, $u=18ms^{-1}$ , $v=0$ and  $a=-7.87ms^{-2}$

$\Rightarrow -18^{2} = 2\times -7.87\times s$

$\Rightarrow s=\frac{18^{2}}{2\times 7.87}=20.58m$

$\Rightarrow s=20.58m$