the first image shows the force acting on the sledge
[tex]\begin{gathered} here\text{ given v}_i=3m\text{/s} \\ v_{f=}12m\text{/s} \\ v_i=velocticy\text{ at the top.} \\ v_f=velocity\text{ at the lower end.} \\ m=10kg \\ h(at\text{ the top\rparen=8m} \\ Hypotenuse(of\text{ the plane\rparen =16m} \\ friction\text{ force = }\mu_kN \\ here\text{ }\mu_k=cofficient\text{ of kinetic friction.} \\ N=Normal\text{ force on the sledge.} \end{gathered}[/tex][tex]\begin{gathered} Work\text{ }done\text{ }by\text{ }all\text{ }theforces=Change\text{ }in\text{ }Kinetic\text{ }Energy\text{ \lparen from work energy thorem\rparen} \\ Wg+W_N+Wf=Kf-Ki \\ WhereWg=work\text{ }done\text{ }by\text{ }gravity. \\ W_N=work\text{ }done\text{ }by\text{ }a\text{ }normal\text{ }force. \\ Wf=work\text{ }done\text{ }byfriction \\ Kf=final\text{ }kinetic\text{ }energy \\ Ki=initial\text{ }kinetic\text{ }energy \\ W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2_i \end{gathered}[/tex][tex]\begin{gathered} force\text{ working on the particle} \\ normal\text{ force N perpendicular to the plane in updard direction.} \\ mgsin\theta\text{ perpendicular to the plane in downward direction } \\ since\text{ particle is in the plane so these forces must be equal.} \\ N=mgcos\theta \\ f=\mu_kN\text{ \lparen friction force acting on the sledge antiparallel to the plane\rparen} \\ mgsin\theta\text{ acting on the particle parallel to the plane.} \\ since\text{ the particle is moving downwards along the plane so the force is along } \\ downward. \\ \end{gathered}[/tex][tex]\begin{gathered} from\text{ Wg + W}_N\text{ + Wf =Kf – Ki} \\ Wg=mgsin\theta *d \\ W_N=0\text{ \lparen angle between N and d is 90 degree\rparen} \\ Wf=-\mu N*d \\ Wf=-\mu mgcos\theta *d \\ so\text{ from} \\ Wg+W_N+Wf=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \\ mgsin\theta *d-\mu mgcos\theta *d=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \\ \lbrace mgsin\theta-\mu mgcos\theta\rbrace *d=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \\ put\text{ all the values} \\ \lbrace10kg*9.8m\text{/s}^2*sin30\degree-\mu(10kg*9.8m\text{/s}^2*cos30\degree)\rbrace16m=\frac{1}{2}*10kg*12^2(m\text{/s\rparen}^2\text{-}\frac{1}{2}*10kg*3^2(m/s\rparen^2 \\ solve\text{ this equation } \\ \lbrace9.8*\frac{1}{2}-\mu(\frac{9.8}{2}*\sqrt{3})\rbrace16=\frac{144}{2}-\frac{9}{2} \\ \lbrace4.9-\mu(8.48)\rbrace16=67.5 \\ 78.4-\mu *135.68=67.5 \\ \mu *135.68=78.4-67.5 \\ \mu=\frac{10.9}{135.68} \\ \mu=0.08 \end{gathered}[/tex]
so μ=0.08 ( dimensionless quantity)
