Question

A cube of mass 'm' slides down the smooth track from height 4R.Then the force track will exert on block at point 2 . Is ______ times the weight of the block.

a. 4
b. 3
c. 2
d. 1

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Kindly give proper solving procedure and full explanation. ​

Answer 1

Solution :-

As no external force is acting upon the block we can say that , the total energy of the system is conserved

By applying law of conservation of energy ,

⇒ KEi + PEi = KEf + PE f

here ,

• KEi = initial kinetic energy
• PEi = initial potential energy
• KE f = final kinetic energy
• PEf = final potential energy

⇒ mv² / 2 + mgh = mv' ² / 2 + mgh'

Initially , the block was at rest ( v = 0 )

⇒ mgh = mv'² / 2 + mgh'

⇒ gh = v'² / 2 + gh'

⇒ 4Rg = v'² /2 + 2Rg

⇒ v'² / 2 = 2Rg

⇒ v'² = 4Rg

We need to find the force exerted by the track on the block at point 2

At point 2 , The track exerts normal force (N) on the block in downward direction along with the gravitational force (mg)

Both these forces provide the necessary centripetal force to the block for it to move in circular motion .

Hence ,

⇒ N + mg = mv'² / r

⇒ N = mv'² / R  - mg

⇒ N = 4Rmg/ R - mg

⇒ N = 4mg - mg

⇒ N = 3mg

The force track will exert on block at point 2  is 3 times the weight of the block.