Question

a) Draw a diagram to show the forces acting on the sledge b) Consider mechanical energy to find the coefficient of friction between the sledge and the ground

Answer 1

the first image shows the force acting on the sledge

`\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}`
`\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=(1)/(2)mv_f^2-(1)/(2)mv^2_i \end{gathered}`
`\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}`

so μ=0.08 ( dimensionless quantity)