Question

The kinetic energy of an object with a mass of `5 kg` constantly changes from `72 J` to `480 J` over `12 s`. What is the impulse on the object at `2 s`?

Answer 1

Assume that the kinetic energy is increasing at a constant rate. After 2s, the impulse on the object would have been `10.58 \quad Kg\cdot m/s`

Explanation:

The impulse exerted on an object equals the change it in its momentum

`Imp= \Delta p= m(v_f-v_i)`

The object's initial kinetic energy is 72 J, so
`72J=1/2 m v_i^2 \quad \quad \implies v_i=5.37m/s`

To find the impulse on the object at 2s we need to find the speed of the object, `v_f`, at 2s.

We are told that the kinetic energy changes constantly. The kinetic energy changes by `(480J-72J=408J)` over 12 seconds.

This means that the kinetic energy changes at a rate of:
`{408J}/{12 s}=34J/s`

In two seconds the kinetic energy will have increased by `34J/s*2s=68J`

Therefore, at 2s the kinetic energy is `(72J+68J)=140J`. This allows us to solve for the `v_f` at 2s

`140J=1/2mv_f^2\quad \quad \implies v_f=7.48 m/s`

Now we have to make sure `v_f` and `v_i` have the right signs when we find `\Delta p`. Assuming the kinetic energy is constantly increasing, `v_f` and `v_i` will be in the same direction and have the same sign.

Substitute `m`, `v_i`, and `v_f` to solve for the impulse.
`Imp=\Delta p=(5 Kg)(7.48m/s-5.37m/s)=10.58\quad Kg\cdot m/s`