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,

$N=mg-(37\times\sin (40^{\circ}))$

The force of friction acting on the sled is,

$F_{}=\mu_k(mg-37\times\sin (40^{\circ}))$

where g is the acceleration due to gravity.

Substituting the known values,

$\begin{gathered} F=0.11(11.5\times9.8-23.78) \\ F=9.79\text{ N} \end{gathered}$

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