Question

Consider someone driving down the highway at 70 miles per hour (31.3 m/s). The total mass of the driver and their car (as well as everything else in the car) is 1600kg.
Part A- How much work must friction do to make the car stop?

Part B- A tree suddenly falls on the highway 55m in front of the car, and the driver just barely manages to stop before hitting the tree. What is the magnitude of the force of friction necessary to do the work you calculated in part A just before the car hits the tree?

Part C- Now imagine that the driver had been traveling at 75 miles per hour (33.5 m/s) at the start instead. The coefficient of friction is still the same, so the force from friction will still be what you calculated in part B, which means the work done by friction will be the same as in part A. Under these circumstances, what will the car's speed be when it hits the tree?

Answer 1

Answer:

A. -783752 J

B. -14250 N

C. 11.94 m/s

Explanation:

A. Work done is given as the change in kinetic energy, hence,

W = ΔKE

W = KE(final) - KE(initial)

KE(final) = 0 since the car comes to rest.

=> W = -KE(initial)

W = -½*m*u²

W = -½ * 1600 * 31.3²

W = -783752J

The negative indicates that the force responsible for the work is acting opposite the direction of the motion.

B. Work done is also given as:

W = F * d

Where F = Force

d = distance moved

=> F = W/d

F = -783752/55

F = -14250 N

The negative indicates that the force acts against the direction of motion.

C. The initial velocity, u, is now 33.5m/s.

Going back to the formula for work done in A,

W = KE(final) - KE(initial)

KE(final) is not zero in this case because the car doesn't come to rest before collision, hence,

W = ½*m*v² - ½*m*u²

Since the force is the same, then, work done is the same,

-783752 = ½*m*v² - ½*1600*33.5²

-783752 = ½*m*v² - 897800

½*m*v² = 897800 - 783752

½*m*v² = 114048

v² = (114048 * 2) / (1600)

v² = 142.56

v = 11.94 m/s