Physics homework help regarding energy and friction

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SUMMARY

The forum discussion centers on a physics homework problem involving a car with a mass of 500 kg and a maximum power output of 75 kW, ascending a hill with an angle of 11.534 degrees at a constant speed of 30 m/s. The energy dissipated by frictional forces and the required frictional force were calculated, revealing a discrepancy when the calculations yielded a negative value for friction. The participants emphasized the importance of recognizing that the kinetic energy remains constant due to the constant speed, thus necessitating a review of the energy balance equation: GPE - (KE + Frictional forces) = 0.

PREREQUISITES
  • Understanding of gravitational potential energy (GPE) and kinetic energy (KE)
  • Familiarity with basic trigonometry and right-angle triangles
  • Knowledge of power calculations in physics
  • Ability to analyze energy conservation in mechanical systems
NEXT STEPS
  • Review the principles of energy conservation in physics
  • Study the relationship between power, work, and energy
  • Learn about the effects of friction on mechanical systems
  • Explore advanced topics in dynamics, including forces on inclined planes
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of energy, friction, and power in mechanical systems.

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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