nope, not right.
final Ke+workdoneonfrciton=force*distance
1/2 m vf^2+870*d=1260*d
1/2 m vf^2=distance*(1260-870)
1/2 *1500*9.5^2)/390=distance
I think that will be the same answer that I gave you before.
A 1500 kg car accelerates from rest under the actions of two forces. One is a forward force of
1260 N provided by traction between the wheels and the road. The other is a 870 N resistive
force due to various frictional forces. Use the work-energy theorem to determine how far the
car must travel for its speed to reach 9.5 m/s.
Ummm don't understand. Is this correct?
1/2[1500][9.5]=[1260-870]cos[d]
1 answer