# Kinetic Energy, Work Problem

## Homework Statement

A car of mass 1240 kg has kinetic energy of 116 kJ. There is a stalled SUV of mass 2710 kg blocking the intersection. Because he obliviously chatting away on his cell phone, the driver of the car plows into the SUV without braking. The two vehicles stick together (a perfectly inelastic collision) and slide to a stop.

## Homework Equations

a) How much work is done against the force of friction, during the time when the vehicles are sliding?

b) The answer to a) above is less than the original kinetic energy. Assume that the energy was transformed to work done by crushing the cars and that the cars were crushed a total of 77.4 cm. What, then, was the magnitude of the average force during the collision?

Assume that the wheels of both cars are locked and do not roll. Also assume that the pavement is dry and that the coefficient of friction between the cars and the road is μ = 0.960.

c) If the pavement was level, how far did the two vehicles slide?

d) Suppose that the pavement was not level, that the car was heading up a hill whose slope was 12.1°. Then the distance the two “shmooshed” vehicles slid uphill was

e) Suppose this time that the car had been heading down the 12.1° grade when it hit the SUV. The distance of slide would then be

## The Attempt at a Solution

a) since perfectly inelastic....

m1*KE1 +m2*KE2 = (m1+m2) * KE
(1240)(116) + (2710)(0) = (1240+2710) KE
KE = 36415 J = 36.415 kJ

b) W = F*d
36415 J = F (0.774 m)
F = 47048.036 N

c) Ffr = u*(mg)
W = Ffr*d
W = u(mg)* d
36415 J = (.960)(92710 kg +1240kg)(9.8) * d
d = 0.9799 m

d) W = Ffr * cos (theta) * d
36415 J = (.960)(92710 kg +1240kg)(9.8) * cos(12.1) * d
d = 0.8433 m

e) W = Ffr * sin (theta) * d
36415 J = (.960)(92710 kg +1240kg)(9.8) * sin(12.1) * d
d = 3.9337 m Still unsure about the formula differences of d and e. ???
Can anybody help me out?? ## Answers and Replies

well off the top of my head, d fails to account for the work done by gravity in coming to a stop, and vice versa. You resolved the frictional force correctly I think but then neglected the work done vs mgh.

am i also missing a factor for part b?

be right back--playing some net gammon.

well the question assumes a perfectly inelastic collision--is this like a perfect storm? Inelastic assumes energy is not conserved, but you set it up as it was.