Question

The curved portions are smooth and horizontal surface is rough. The block is released from $P$. At what distance from $A$ it will stop (if $\mu =0.2$)

Answer 1

During the fall from 1m height,the block will gain kinetic energy and the total energy will be purely kinetic on coming to point A,which was purely potential energy while sitting on top of the curve.

Applying law of conservation of energy,

we can say, $mgh=\frac{1}{2}m{v}^2$ (where, v is the velocity at A)

So, $v^2=19.62$

Now,the frictional force that will be acting on the block is $\mu N=\mu mg$ ($\mu$ = coefficient of friction$=0.2$)

So,deceleration while going along the horizontal part will be $\mu mgm=\mu g$

So,if it stops by going distance x then applying,

$v^2=u^2-2ax$

we get,$0^2={19.62}^2-2\times 1.962x$

So,$x=5m$

But,given length of the horizontal part is $2m$, so let us see what will be the velocity,after traveling through $2m$,

So, $v^{\prime 2}={19.6}^2-2\times 1.962\times 2=11.76$

So,at point B total energy = kinetic energy

Now,if it goes up by height $s$ then, again applying law of conservation of energy,

$\frac{1}{2}m{v}^{\prime 2}=mgs$

So, $s=0.6m$

Now,after reaching that point it will again come down and gain the same velocity at point B while it was going up,so now if it moves distance r towards A, we can check like previously,

$0^2=11.76-2\times 1.962\times r$

So, $r=3$

So,it will still not come to rest still now, so lets find velocity after moving$2m$ to reach at A

So,after rising up when coming down,it will have same velocity at A,

So,this time if it stops after going y ,we can say,

$0^2=4-2\times 1.962\times y$

, $y=1.01$

Almost about 1m from A after these above set of rides the block will stop.