The kinetic energy of an object with a mass of 5 kg5kg constantly changes from 72 J72J to 480 J480J over 12 s12s. What is the impulse on the object at 2 s2s?

1 Answer
Apr 19, 2016

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