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?

A block of mass 1 kg is given horizontal velocity `v_(0)=10sqrt(3)(m)

1.	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» Correct Answer - 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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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?

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