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#?

1 Answer
Apr 19, 2016

Answer:

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#