Answer:
[tex]\mu mgd[/tex]
Explanation:
In the situation described in the problem, there are 3 forces acting on the coffee cup:
- The weight of the cup, acting downward
- The normal force exerted by the horizontal dash on the cup, acting upward
- The force of friction, acting forward and "keeping" the cup travelling together with the car, without sliding backward
For the purpose of this problem, we ignore the weight and the normal force, since they cancel each other out.
The force of friction here pushes the cup forward, allowing it to move together with the dash and the car, without sliding. The magnitude of this force of friction is
[tex]F_f = \mu mg[/tex]
where
[tex]\mu[/tex] is the coefficient of static friction between the cup and the dash
m is the mass of the cup
g is the acceleration of gravity
Since the cup is moving forward together with the car, it means that it has a certain displacement [tex]d[/tex], and therefore the force of friction is performing work, equal to the product between force and displacement, so:
[tex]W=F_f \cdot d = \mu mg d[/tex]
