Aperson places a cup of coffee on the roof of her car while shedashes back into the house for a forgotten item. when she returnsto the car, she hops in and takes off with the coffee cup still onthe roof.
(a) if the coefficient of static frictionbetween the coffee cup and the roof of the car is 0.22, what is the maximum acceleration the car canhave without causing the cup to slide? ignore the effects of airresistance.
1 m/s2
(b) what is the smallest amount of time in which the person canaccelerate the car from rest to 13 m/s andstill keep the coffee cup on the roof?

(a)

Along the vertical direction, the acceleration of the cup is zero, therefore the equation of the forces is:

where N is the normal reaction and (mg) is the weight of the cup, with m being its mass and g the acceleration of gravity. From this equation we get

(1)

Along the horizontal direction, the only force acting is the force of friction. The maximum force of friction acting on the cup is

where is the coefficient of static friction between the cup and the roof of the car. The maximum acceleration that the car can substain withouth the cup sliding away corresponds to the maximum force of friction on the cup, so we can write:

And solving for and using eq.(1), we get:

(b) 5.9 s

To solve this part of the problem, we can use the following SUVAT equation:

where

v is the final velocity

u is the initial velocity

a is the maximum acceleration

t is the smallest amount of time

In this problem,

v = 13 m/s (final velocity of the car)

u = 0 (initial velocity)

a = 2.2 m/s^2

Solving for t,

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.