Physics homework help regarding energy and friction

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Homework Help Overview

The discussion revolves around a physics problem involving energy, friction, and the dynamics of a car moving up a hill. The original poster presents a scenario where a car of mass 500 kg has a maximum power output of 75 kW and is traveling at a constant speed of 30 m/s up an incline of 11.534 degrees. The goal is to determine the energy dissipated by frictional forces and the corresponding frictional force.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the energy gained from gravitational potential energy and kinetic energy, while also considering frictional forces. They express confusion over obtaining a negative value for frictional force. Some participants question the assumptions made regarding constant speed and the relationship between vertical and diagonal velocities.

Discussion Status

The discussion is ongoing, with participants providing guidance on reviewing assumptions and definitions related to the problem. There is an acknowledgment of the original poster's calculations, but also a suggestion to reconsider the context of kinetic energy in relation to the problem's parameters.

Contextual Notes

Participants note the importance of the assumption of constant speed and its implications for the calculations being made. There is a focus on ensuring that the definitions used in the equations align with the scenario presented.

Dan_321
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Homework Statement


The maximum power output of a car of mass 500kg is 75kW. Up a hill of angle 11.534, its maximum (constant) speed achievable with this power is 30m/s.

What is the energy dissipated by frictional forces every second, and what must the frictional force be?

Homework Equations


mgh=loss/gain of GPE, KE=(mv^2)/2=Kinetic energy gained/lost

The Attempt at a Solution


As this is a right angle triangle, we know that 30sin(11.534) will equal the vertical velocity going up (approx 6m/s). From there we know what mgh is: 6m/s*10*500=30000J gained per second. Since GPE-(KE+Frictional forces) must equal 0, 30000-((500*30^2)/2+F) should give F, but it turns out to be negative. How does this work?
 
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Dan_321 said:
its maximum (constant) speed

The assumption is that the speed is constant. Review the terms in your calculation given that information.
 
RPinPA said:
The assumption is that the speed is constant. Review the terms in your calculation given that information.
Yes, but surely the vertical velocity is going to be different from the diagonal velocity?
 
Yes. I agree with your calculation that the vertical velocity is a constant 6 m/s. Now look at your definitions in "relevant equations".
 
Dan_321 said:
GPE-(KE+Frictional forces)
Why, in this context? As @RPinPA notes, there is no change in KE.
 

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