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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

A 395 kg sled is pulled forwardwith a 728 N force which causesan acceleration of 0521 msHow much friction force actson the sledUnit NEnterUsing Alternative class=

Respuesta :

We have

Where

[tex]F-FR=ma[/tex]

F=72.8 N

m=39.5kg

a=0.521 m/s^2

we substitute the values

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

The friction force is 52.22N

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