You are driving in a car with your 0.1 kg coffee cup sitting on the dashboard. T

Question

You are driving in a car with your 0.1 kg coffee cup sitting on the dashboard. The dashboard is sloped at 15 degrees with respect to the horizontal, with the lower part of the dashboard facing the rear of the car. The static coefficient of friction between the cup and the dashboard is 0.7 and the kinetic coefficient of friction is 0.5.

Part A - What is the fastest that you can accelerate without the coffee cup sliding on the dashboard?

Part B - What is the maximum rate that you can slow down while braking? (Give just the magnitude of this acceleration.)

Part C - If the coffee cup has coffee in it, it effectively has more mass. With more mass, will these accelerations be larger, smaller, or the same?

Explanation / Answer

let the acceleraion of the car be a

so there will be force m * a on the cup due to this acceleration

force equation of the cup will be

mg * sin(theta) - k * mg * cos(theta) + ma * cos(theta) = 0

net acceleration of the cup is 0 since its in rest

k will be coefficient of static firction

0.1 * 9.8 * sin(15) - 0.7 * 0.1 * 9.8 * cos(15) + 0.1 * a * cos(15) = 0

a = 4.2341 m/s^2

a) fastest one can accelerate car = 4.2341 m/s^2

in case of breaking m * a anf friction force will act in opposite direction so

mg * sin(theta) + k * mg * cos(theta) - ma * cos(theta) = 0

0.1 * 9.8 * sin(15) + 0.7 * 0.1 * 9.8 * cos(15) - 0.1 * a * cos(15) = 0

a = 9.4859 m/s^2

b) maximum rate one can slow down = 9.4859 m/s^2

c) if the coffee cup has more mass then also the value of accelerations will remain same as

mg * sin(theta) + k * mg * cos(theta) - ma * cos(theta) = 0

g * sin(theta) + k * g * cos(theta) - a * cos(theta) = 0

final equation is independent of mass

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