Does Static Friction Contribute to Work in an Inclined Car?

AI Thread Summary
The discussion centers on the work done on a car descending an incline with constant acceleration, specifically questioning the role of static friction. It is established that static friction does not perform work since both the normal force and static friction have zero displacement. However, there is confusion regarding a grading error where a response indicating zero work was marked wrong. The clarification suggests that the calculation should consider the displacement of the car's center of mass rather than the point of contact, leading to the concept of pseudowork. Ultimately, the thread highlights the nuances in understanding work in the context of forces acting on a moving vehicle.
Zhalfirin88
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Homework Statement


A car is going down an incline with a constant acceleration.
How much work was done on the car by the force of the road going down the hill? (neglect energy losses due to air resistance, rolling friction, etc.)

The Attempt at a Solution


There are 2 forces acting on the car from the road, the normal force and static friction. But the displacement of both the normal force and static friction are both 0, so how is static friction doing work?
 
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Zhalfirin88 said:
There are 2 forces acting on the car from the road, the normal force and static friction. But the displacement of both the normal force and static friction are both 0, so how is static friction doing work?
Technically, the static friction does no work on the car.
 
That's exactly what I put, 0, and it was marked wrong. I'm trying to figure out why it was marked wrong
 
Work done, ΔW= F.Δx
Because Ff opposes the motion, it is directed opposite to x

the work done on the car by friction,
W=-Ff.x
 
Last edited:
Zhalfirin88 said:
That's exactly what I put, 0, and it was marked wrong. I'm trying to figure out why it was marked wrong
They probably wanted you to calculate F*Δx, where Δx is the displacement of the center of mass, not the point of contact. Strictly speaking, that's not work, but pseudowork.
 
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