kinetic energy, frictional work problem?

Question

A toy car is sliding with a kinetic energy of 3.9 J when it locks up its tires and slides for 74 cm, while a frictional force of 0.74 N acts against it. What is the kinetic energy (in Joules) of the car after this frictional work is done?

I'd appreciate it if anyone could offer some assitance.

Answer 1

Work is equal to a force times a distance. So you can calculate how much work was done by the frictional force using the first equation in the link below.

W = 0.55 joule

So this is how much of the kinetic energy of the toy car was used up in skidding the wheels. So now all you need to do is calculate how much kinetic energy is left over.

KE = 3.35 joule

Answer 2

Total energy is constant for the car; so TE = PE + KE + WE; where PE is potential energy, KE is kinetic energy, and WE is work energy. TE will remain fixed because of the conservation of total energy law. What changes is the contribution of each term...KE, PE, and WE.

OK, then, sliding before braking TE(0) = KE = 1/2 mv^2 = 3.9 kg (m/sec)^2; since no work is done and there is no potential energy (mgH) as H = 0, the height above ground level.

After sliding and braking for d = .74 meters, total energy is TE(B) = KE(B) + WE(B); there still is no PE, but now we have work against the friction force WE(B) = Fd = .74*.74 kg (m/sec)^2 ~ .5 Joules. (You can do the math for more precision.)

From the conservation of energy TE(0) = 3.9 = KE(B) + F*d = KE(B) + .5 = TE(B). Thus, KE(B) = 3.9 - .5 ~ 3.4 Joules (approximiately) of kinetic energy remaining.