A person places a cup of coffee on the roof of her car while she dashes back int

Question

A person places a cup of coffee on the roof of her car while she dashes back into the house for a forgotten item. When she returns to the car, she hops in and takes off with the coffee cup still on the roof.

1) If the coefficient of static friction between the coffee cup and the roof of the car is 0.23, what is the maximum acceleration the car can have without causing the cup to slide? Ignore the effects of air resistance.

2) What is the smallest amount of time in which the person can accelerate the car from rest to 17 m/s and still keep the coffee cup on the roof?

Express both answers in two significent figures.

Explanation / Answer

Forces acting on the coffee cup:
Weight (m*g): downward
Normal force from the roof (N): upward
Friction force from the roof (F): forward

Vertical force balance:
N = m*g

Newton's second law for the horizontal force of friction:
F = m*a

Recall formula for maximum deliverable force of friction:
F = mu_s*N = mu_s*m*g

Thus:
a = mu_s*g

And for the time to reach 22 m/s:
a = (vf - vi)/t

solve for t, knowing vi =0:
t = vf/a

Substitute:
t = vf/(mu_s*g)

Summary:
a = mu_s*g
t = vf/(mu_s*g)

Data:
mu_s:=0.23; g:=9.8 N/kg; vf:=17 m/s;

Results:
1. a = 2.254 meters/second^2
2. t = 7.542 seconds

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