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

Answer

2

Explanation

`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`