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My understanding of work is that it is the product of a force and the displacement of the object in the direction of the force. So if an object is already moving at 1m/s and we apply a force of 1N in the same direction of motion, is the work done 1J after the object has moved 1m?

This seems wrong to me as it feels like the work done should be less than the case where the object starts with 0 velocity.

user40990
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  • An external force "adds" a kinetic energy according to the Newton equation, which is unambiguous way to proceed. Try to solve it and analyse the result as a function of the initial velocity and the given distance. – Vladimir Kalitvianski Jun 16 '18 at 18:17

1 Answers1

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W = F(avg)$\Delta$x

If you look at the equation for work you can see that what you should already be correct. In high school level problems in work & energy, you can simply find the work of friction (same idea) by multiplying the friction force by the distance the friction force acted on the object.

Hope this helps!

Chawit K
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