A block of mass 5 kg slides down a rough...

A block of mass 5 kg slides down a rough inclined surface. The angle of inclination is `45^(@)`. The coefficient of sliding friction is 0.20. When the block slides 10 cm, the work done on the block by force of friction is

`-(1)/(sqrt2)J`

`-sqrt2J`

`-1J`

Text Solution

Verified by Experts

The correct Answer is:

(a) Work done by frictional force, `W_(f)=fs cos 180^(@)`
`=(mu mg cos theta)(-1)(s)`
`=-0.2xx5xx10xx(1)/(sqrt2)xx0.1=-(1)/(sqrt2)J`

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