How to compute Work Done in this case?

Asked by  | 25th Sep, 2008, 09:53: PM

Expert Answer:

Friction,f=(coefficient of kinetic friction)mg=0.2(20)=4 N

Acceleration,a=(F-f)/m=(7-4)/2=1.5 ms-2

Distance moved,s=1/2(1.5)(100)=75 m



Work done by applied force=(7)(75)=525 N

Work done by friction=(4)(75)=300 N

Work done by applied force=(3)(75)=225 N

Change in kinetic energy =Work done by applied force=(7)(75)=525 N

Answered by  | 27th Sep, 2008, 01:21: PM

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