Solving the Mystery of an Electrical Toy Car's Power

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Homework Help Overview

The problem involves an electrical toy car moving up an inclined plane at a constant speed, with a focus on calculating the average power exerted by the car. The scenario includes considerations of mass, friction, and gravitational forces, specifically in the context of a past exam question.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the interpretation of friction in the context of the problem, questioning whether it refers to retarding forces or the force propelling the car. There is discussion about the implications of the car's motion and the nature of the question itself.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications regarding the interpretation of friction and the mechanics involved. There is no explicit consensus, but some guidance has been offered regarding the nature of the forces acting on the car.

Contextual Notes

Participants note potential ambiguities in the problem statement, including the definition of friction and its role in the car's motion. There is a suggestion that the question may be poorly constructed, leading to confusion about the expected answer.

tsw99
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Homework Statement



I encountered this past exam paper question in Hong Kong when I was giving tutorial to a high school student. I am confused with the answer provided by the official authority. The question is (in exact wordings):

An electrical toy car of mass m goes up an inclined plane of inclination 30° with constant speed v. The friction acting on the car is half of the weight of the car. What is the average power of the car?
A. 1/2 mgv B. mgv C. 3/2 mgv D. 2mgv

Homework Equations


The Attempt at a Solution


The friction f provides by the ground to propel the car is exactly canceled by the component of weight down the inclined plane, hence the power is fv which is equal to mg(sin30°)v=1/2 mgv which is A. However the official answer is B. I have no idea how to arrive at the answer B and I believe the authority must have their own grounds. I even think that B is a "trap" choice when I first read this question.
Any help is appreciated. Thanks.
 
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It sounds like "the friction acting on the car" here refers to friction that is retarding the motion of the car, and not the friction you mention that is propelling the car forward.

Hope that helps.
 
Redbelly98 said:
It sounds like "the friction acting on the car" here refers to friction that is retarding the motion of the car, and not the friction you mention that is propelling the car forward.

Hope that helps.

Yours is a probable answer. Does that mean the car is skidding up the incline? So the friction does not provide the force to travel upward?
Frankly speaking, I never saw an electric toy car travels in this way. I believe it is a poorly set question.
 
Generally speaking, contact friction, such as between the tire and the ground is not considered to do work.

The friction acting to impede the car as it goes up the ramp would likely be from the non-driven wheel-axle system and perhaps some piece of the car dragging on the ground.
 
tsw99 said:
Yours is a probable answer. Does that mean the car is skidding up the incline? So the friction does not provide the force to travel upward?
Frankly speaking, I never saw an electric toy car travels in this way. I believe it is a poorly set question.
SammyS provided a good explanation, so I'll just add that no, I did not mean the car is skidding. Friction between the wheel and ground does provide the force to propel the car, however that is not the friction being referred to in the problem statement.
 

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