# The force f acting on a body varies with its displacement x as f=k×x^(-2/3). The power delivered by the force will be proportional to what?

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The work delivered by force equals `W = f*x` , when x is the displacement. However, when F is not constant with respect to x, the work cannot be calculated as the product of the force and displacement. However, this relationship is still true for a work `DeltaW` done on a very small displacement `Deltax` , because for a very small change in x, the force is approximately constant:

`DeltaW = fDeltax` , or in differential notation, `dW = Fdx`

The total work would then be the sum of the work done on small displacements, or the integral:

`W = int fdx` .

The given force is `f = kx^(-2/3)` , so the work would be

`W = int_0 ^x kx^(-2/3) dx = 3kx^(1/3)`

The power delivered is the rate at which the work is done:

`P = (dW)/(dt) = (d(dkx^(-1/3)))/(dt) = kx^(-2/3) (dx)/(dt)` , where dx/dt is the velocity.

**The power will be proportional to x^(-2/3) and v = dx/dt.**

The power delivered by the force can be calculated as the ratio of work done per unit time. That is:

power = work done/time

Work done by the force is the product of force and the displacement. That is,

Work done = force x displacement.

Hence, power = force x displacement/time

The force is a function of displacement (x) and is given as:

`f = kx^(-2/3)`

Thus, the power generated by the force will be given as:

`power = fx/t = (kx^(-2/3)x)/t = (kx^(1/3))/t`

Thus, the power generated by the force will be directly proportional to the displacement (raised to the power 1/3) and is inversely proportional to time t. In other words,

`Power alpha x^(1/3)`

`Power alpha 1/t`

Thus, given the displacement and the time taken, we can calculate the power generated by a certain force on a body.

Hope this helps.