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What can we say about the work of a force applied to a point with a velocity $v=0$?

My guess is $\delta W = P(t) dt \implies W = \int F\cdot v \ dt$ and then since $v=0$ we have that $W=0$.

I am not sure if it is "$dt$" in the integral or something else.

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  • $\begingroup$ The question is ambiguous which has lead to two conflicting answers. It is not clear whether the word "force" is meant to apply to a single force on an object subject to several forces, or to the net force on the object. $\endgroup$
    – garyp
    Jan 6, 2021 at 12:51
  • $\begingroup$ A single one, sorry if it was not clear $\endgroup$
    – Kilkik
    Jan 6, 2021 at 13:27

3 Answers 3

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That is correct. A force on an object at rest does no work.

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Work is defined as dot product of force and displacement. For a differential time dt, when velocity is 0 the object does not undergo displacement and hence it is true that the work on the object in that differential time is zero. Although it will change soon after the as the object will start moving under the effects of the force. It is the same as multiplying any number with 0 will give you zero. A force that does not produce a change does not spend energy.

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I dont agree with @Dale This is only possible when there is some external agent which is holding the object resulting in no motion

But if the force acts there is some accelerate causing the body to gain some velocity

Your function v is a variable which can be written as


dv/dt=F/m

V=Ft/m


If you say that the force isn't sufficient to cause a displacement like it's less than maximum friction and object has 0 velocity forever then its work done is definitely 0

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  • $\begingroup$ It does not matter how comes that the object is at rest. If it is, the work is zero. And your last formula works only for constant force. $\endgroup$
    – nasu
    Jan 6, 2021 at 13:42
  • $\begingroup$ @nasu it's nowhere mentioned force is variable. Moreover even if its variable you can integrate in same way resulting in different result $\endgroup$
    – imposter
    Jan 6, 2021 at 14:14
  • $\begingroup$ U mean if a box is at rest initially and a force is being applied to the box for a dt time resulting in displacement and ultimate ly a velocity then the force is doing 0 work. $\endgroup$
    – imposter
    Jan 6, 2021 at 14:16
  • $\begingroup$ No, i mean as long as velocity is zero the work is zero. What happens later is another question. $\endgroup$
    – nasu
    Jan 6, 2021 at 14:42

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