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A block of mass 1kg is given horizontal ...

A block of mass `1kg` is given horizontal velocity `v_(0)=10sqrt(3)m//s` from origin along `x`-axis on rough horizontal surface where coefficient of friction is `mu=mu_(0)(1+x/10)`, where `mu_(0)=1`. It is found that maximum power loss due to frictioni is `100K` in S.I. units. What is value of `K`?

Text Solution

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The correct Answer is:
2
`v(dv)/(dx)=-mu_(0)g(1+(x)/(10))`
`impliesv=sqrt(300-20x-x^(2))`
power loss
`=(1+(x)/(10))mgsqrt(300-20x-x^(2))`
`impliesP_(max)=200`
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