PHYSICS EXAM REVIEW PROBLEM-Exam Friday, please help!

Question

You are driving home with a 5.5 kg birthday cake sitting on your front seat, which is level. The coefficient of static friction between the cake and the seat is 0.40 and the coefficient of kinetic friction 0.25. What is the maximum acceleration during braking that your car can have if the cake is going to stay on the seat?

Answer 1

Draw a free-body diagram:

The sum of the forces in the y-direction (∑F_y) must be zero because there is no acceleration, or movement at all, in the y-direction (the motion occurs in the x-direction).

So (by looking at the free-body diagram): ∑F_y = F_N - W = 0 (where F_N is the normal force, or the force that the seat is pushing up against the cake with and "W" is the weight of the cake).

The weight of the cake is its mass multiplied by the acceleration due to gravity ("g", which is 9.8 m/s₂)

So, since F_N - W = 0 then:

F_N - mg = 0

F_N = mg

F_N = (5.5)(9.8) = 53.9 N

The force of friction (F_F) is proportional to the normal force:

Where F_F = μF_N (where μ is the coefficient of friction, in this case since the cake isn't sliding yet, you would use μ = 0.40)

So F_F = (0.4)(53.9) = 21.56 N

This is the maximum force that friction can provide to you to keep the cake from moving.

Newton's 2nd Law says F = ma so:

21.56 = (5.5)a

a = 21.56/5.5 = 3.92 m/s²

If there is any more acceleration than that, the force of friction will not hold the cake in place.

It would be appropriate to round the answer to 2 significant figures so a = 3.9 m/s² also since the car is decelerating, you could call the answer a = -3.9 m/s²