The kinetic energy of an object with a mass of #1 kg# constantly changes from #243 J# to #658 J# over #9 s#. What is the impulse on the object at #3 s#?

1 Answer
Apr 30, 2016

Answer:

You must aknowledge that the key words are "constantly changes". Afterwards, use the kinetic energy and impulse definitions.

Answer is:

#J=5.57# #kg*m/s#

Explanation:

The impulse is equal to the change of momentum:

#J=Δp=m*u_2-m*u_1#

However, we are missing the velocities.

Constantly changing means that it changes "steadily". This way, we can assume that the rate of change of the kinetic energy #K# with respect to time is constant:

#(ΔK)/(Δt)=(658-243)/9=46.1 J/s#

So for every second the object gains #46.1# joules. For three seconds:

#46.1*3=138.3# #J#

Therefore the kinetic energy at #3s# is equal to the initial plus the change:

#K_(3s)=K_(i)+K_(ch)=243+138.3=381.3# #J#

Now that both kinetic energies are known, their velocities can be found:

#K=1/2*m*u^2#

#u=sqrt((2K)/m)#

#u_1=sqrt((2*243)/1)=22.05m/s#

#u_2=sqrt((2*381.3)/1)=27.62m/s#

Finally, the impulse can be calculated:

#J=Δp=m*u_2-m*u_1=1*27.62-1*22.05#

#J=5.57# #kg*m/s#