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A 60-kg box is being pushed a distance of 7.9 m across the floor by a force Upper POverscript right-arrow EndScripts ⁢ whose magnitude is 194 N. The force Upper POverscript right-arrow EndScripts ⁢ is parallel to the displacement of the box. The coefficient of kinetic friction is 0.23. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.

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Answer:

Step-by-step explanation:

Given

mass of box=60 kg

distance=7.9 m


\vec{P}=194 N

Force is parallel to displacement

coefficient of kinetic energy
\mu =0.23

Work done by Force


W=P\cdot x=Px\cos \theta


W=Px\cos 0


W=194* 7.9=1532.6 N

Work done by Friction force


W=f_r\cdot x


W=\mu Nx \cos 180


W=-0.23* 60* 9.8* 7.9=-1068.396 J

negative sign indicates that force is opposite to displacement

Work due to Weight and Normal reaction is zero as the force is perpendicular to Displacement

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