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A block is placed on the top of a plane ...

A block is placed on the top of a plane inclined at `37^@` with horizontal. The length of the plane is `5m`. The block slides down the plane and reaches the bottom.

a. Find the speed of the block at the bottom if the inclined plane is smooth.
b. Find the speed of the block at the bottom if the coefficient of friction is `0.25`.

Text Solution

Verified by Experts

Let h be the height of the inclined plane.
`h=5sin37^@=3m` ltbgt a. As the block slides down the inclined plane, it loses gravitational potential energy and gains KE.
Loss in GPE=gain in KE
mg (loss in height) `=KE_j-KE_i`
`implies mgh=1/2mv^2-0`
`impliesv=sqrt(2gh)=sqrt(2xx9.8xx3)=7.67ms^-1`
b. As the blocks comes down, it loses GPE. It gains KE and does work against friction.
Loss in GPE=gain in KE+work done against friction
`=mgh=(1/2mv^2-0)+(mumgcos37^@)s`
`=3mg=1/2mv^2+(0.25)xxmgxx4/5xx5`
`impliesv=sqrt(4g)=6.26ms^-1`
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