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?

• A. $1m$
• B. $2m$
• C. $3m$
• D. $4m$

Answer 1

The correct option is A $1m$

Given: the curved portions are smooth and horizontal surface is rough. The block is released from P

To find at what distance from A it will stop

Solution:

Applying the law of conservation of energy

$mgh=\frac{1}{2}m{v}^2⟹v^2=2gh⟹v^2=19.62$

Frictional force is

$=\mu N=\mu mg⟹a=\frac{\mu mg}{m}⟹a=\mu g=0.2\times 9.8=-1.962m/s^2$

It stops by going distance x, therefore

$v^2=u^2-2ax⟹19.62=0^2-2(-1.962)\left(x\right)⟹x=5m$

But path is of 2m

Therefore $v^{\prime 2}=19.62-(2\times 1.962\times 2)=11.77$

$KE=\frac{1}{2}m{v}^{\prime 2}⟹\frac{1}{2}m{v}^{\prime 2}=mgs⟹S=\frac{v^{\prime 2}}{2g}=\frac{11.77}{2\times 9.8}⟹S=0.6m$

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 then

$0^2=11.77-(2\times 1.962)r⟹r=3$

So, it will still not come to rest, so lets find the velocity after moving 2m to reach

$u^2=11.77-(2\times 1.962\times 2)=4$

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

$0^2=4-(2\times 1.962\times y)⟹y=1$

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

is the distance from A at which it stops.