A block of mass `m` is given velocity `v_0` on a rough horizontal surface. After traveling a distance d. Its speed becomes `(v_0)//(2)`. Find the work done against friction. The work donw by friction and the coefficient of friction.

Explanation

`K_1=(1)/(2)mv_0^2`,`U_(g1)=0`
`K_2=(1)/(2)m((v_0)/(2))^2`,`U_(g2)=0`
`K_1+U_1+K_2+U_2+W_f`
`(1)/(2)mv_0^2=(1)/(2)m((v_0)/(2))^2+W_f`
Work done against friction `W_f=(3)/(8)mv_0^2`
Work done by friction `W_f^(')=-W_f=-(3)/(8)mv_0^2`
`W_f^(')=-muNd=-mumgd=-(3)/(8)mv_0^2`
`mu=(3v_0^2)/(8gd)`
Loss in mechanical energy `=` work done against friction
`(1)/(2)mv_0^2-(1)/(2)m((v_0)/(2))=W_f`
`W_f=(3)/(8)mv_0^2`