A classical mechanics problem by James Guevara

A horizontal force of 15 N 15\text{ N} is required to maintain a velocity of 2 m/s 2\text{ m/s} for a box of mass 10 kg 10\text{ kg} sliding over a rough horizontal surface.

What is the total work done on the box in 1 minute?


The answer is 0.

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1 solution

James Guevara
Nov 20, 2016

W t = W f + W f k Wt= Wf+ Wfk

W f = F ( s ) Wf=F(s)

W f = 15 N ( 2 m / s ) ( 60 s ) = 1800 J Wf=15N(2m/s)(60s)=1800J

W f k = F ( s ) c o s 180 Wfk=F(s)cos180

W f k = 15 N ( 2 m / s ) ( 60 s ) ( 1 ) = 1800 J Wfk=15N(2m/s)(60s)(-1)=-1800J

W t = W f + W f k Wt=Wf+Wfk

W t = 0 Wt=0

Verbally, it can be stated as: Since there is no net force acting on the body, by definition of work, which is the dot product of force and displacement, the net work done is zero as well.

Nihar Mahajan - 4 years, 6 months ago

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