Question

A 39.5 kg sled is pulled forwardwith a 72.8 N force, which causesan acceleration of 0.521 m/s?.How much friction force actson the sled?(Unit = N)EnterUsing Alternative

Answer 1

We have

Where

`F-FR=ma`

F=72.8 N

m=39.5kg

a=0.521 m/s^2

we substitute the values

`\begin{gathered} FR=F-ma \\ FR=72.8-(39.5\cdot0.521) \\ FR=52.22N \end{gathered}`

The friction force is 52.22N