Consider a particle raised to height $h$ by applying a constant force of $F$ vertically opposite to the direction of force of gravity on it. The potential energy of the particle is

$$W = mgh$$

But, according to the definition of work, the work done on the particle is

$$W' = Fh$$

According to conservation of energy, $W = W'$. But what if $F>mg$?


1 Answer 1


Good question, and the answer is that $W' > W$ if $F > mg$.

The reason for this is that if $F = mg$ then the net force is zero so the particle travels at a constant velocity. That means it's kinetic energy hasn't changed so the only change is the potential energy.

However if $F > mg$ then the net force is positive and the particle is accelerating upwards. That means when it reaches the height $h$ both the potential energy, $mgh$, and the kinetic energy, $0.5mv^2$, have increased. The difference between $W$ and $W'$ is the extra kinetic energy of the particle.


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