Amount of work and size of force?

Question

A 2.00*10^3kg car has a speed of 12.0m/s. The car then hits a tree. The tree doesn't move and the car comes to rest.
a) What is the amount of work done in pushing the front of the car?
b) Find the sizeof the force that pushed the front of the car in50.0cm

Answer 1

The amount of work = W = Fd = (d(mv)/dt)d; where d is the distance of the crumple measured from first impact through point of rest and F = Ma = the force applied to the front of the car as it traveled d distance and came to a halt.

a The amount of work is equal to the dissipated (converted) kinetic energy at impact = KE = 1/2 Mv^2; where M = 2X10^3 kg and v = 12 m/sec. (You can do the math.)

b. The force = F = W/d = KE/d; where d = .5 m and you've already solved for KE = W. (You can do the math.)

Physics lesson learned: Kinetic Energy dissipated is equal to the work (Fd) expended in moving some mass (m) a distance d. Remember, energy is neither created nor destroyed; so that KE had to go somewhere as the car came to a screeching halt...it went into work.

Answer 2

a) That's the kinetic energy of the car before impact
b) Work is force times distance