A block of mass m starts at rest at heig...

A block of mass m starts at rest at height on a frictionless inclined plane. The block slides down the plane, travels across a rough horizontal surface with coefficient of kinetic friction `mu` and compresses a spring with force constant `k` a distance `x` before momentarilly coming to rest. Then the spring extends and the block travels back across the rough surface, sliding up the plane. The block travels a total distance d on rough horizontal surface. The correct experssion for the maximum height h' that the block reaches on its return is :

`mgh'=mgh-mumgd`

`mgh'=mgh+mumgd`

`mgh'=mgh+mumgd+kx^(2)`

`mgh'=mgh-mumgd-kx^(2)`

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The correct Answer is:

Final `P.E.` of block `=` Initial `P.E.` of block `+` work done by friction
`:. Mgh' =mgh-mu mg d`

A block of mass m slides from rest at a height H on a frictionless inclined plane as shown in the figure. It travels a distance d across a rough horizontal surface with coefficient of kinetic friction mu , and compresses a spring of spring k by a distance x before coming to rest momentarily. Then the spring extends and the block travels back attaining a final height of h. Then

`h = H-2 mu (d + x)`

`h = H + 2 mu (d + x)`

`h = H-2 mud + kx^(2)//mg `

`h= H - 2 mu (d + x) + kx^(2)// 2mg `

sin `theta`

cos `theta`

tan `theta`

0.25

0.275

0.325

0.346