Question

An 11.5 kg sled is pulled with a 37.0 N forceat a 40.0° angle, across ground whereMK = 0.110.What is the force of friction on the sled?

Answer 1

The value of the friction coefficient is,

`\mu_k=0.11`

The mass of the sled is,

`m=11.5\text{ kg}`

The normal force acting on the sled is,

`N=mg-(37*\sin (40^(\circ)))`

The force of friction acting on the sled is,

`F_{}=\mu_k(mg-37*\sin (40^(\circ)))`

where g is the acceleration due to gravity.

Substituting the known values,

`\begin{gathered} F=0.11(11.5*9.8-23.78) \\ F=9.79\text{ N} \end{gathered}`

Thus, the force of friction on the sled is 9.79 N in the opposite direction of the pull.