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#?

Question

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#?

Charles Aeschliman · Accepted 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# The change in momentum equals the impulse: #J=Δp=m*u_2-m*u_1# The velocities are absent, though. 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# Given the two kinetic energies, their velocities can now be determined: #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# At last, it is possible to compute the impulse: #J=Δp=m*u_2-m*u_1=1*27.62-1*22.05# #J=5.57# #kg*m/s#

Charles Aeschliman · Answer

undefined To find the impulse at 3 seconds, you can use the impulse-momentum theorem, which states that impulse equals the change in momentum. First, find the initial and final momentum of the object at 3 seconds using the given kinetic energy values and mass. Then, subtract the initial momentum from the final momentum to find the impulse.