A man of mass `60kg` sitting on ice pushes a block of mass `12kg` on ice horizontally with a speed of `5ms^(-1)` The coefficient of friction between the man and ice and between block and ice is `0.2 If g =10ms^(2)` the distance beteen man and the block when they come to rest is .

Options

A. `6m`
B. `6.5m`
C. `3m`
D. `7m`

Answer

B

Explanation

`S_(1) = (v^(2))/(2mug)`, according to law of conservation of momentum find `v^(1),S_(2) =(v^(1^(2)))/(2mug),S=S_(1)+S_(2)` .