Supposed to be an easy question about WORK

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Homework Help Overview

The discussion revolves around a physics problem involving work, energy, and friction as Jim rides his skateboard down a quarter circular ramp. The problem includes calculations related to gravitational potential energy and kinetic energy, as well as the work done by friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy and kinetic energy at different points on the ramp. Questions arise about the relevance of the ramp's radius and the nature of the frictional force. Some participants suggest alternative approaches to calculate work done against friction.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some guidance has been offered regarding energy conservation and the calculation of work, but there is no explicit consensus on the correct approach or solution.

Contextual Notes

Participants express confusion about the problem setup, particularly regarding the implications of the ramp's curvature and the calculations involved. There is a mention of potential constraints related to homework rules and the need for clarity on definitions of energy types.

mms05
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A problem about work

Hi! I need some help with this question:

Jim rides his skateboard down a ramp that is in the shape of a quarter circle with a radius of 5.00 meters. At the bottom of the ramp, Jim is moving at 9.00 m/s. Jim and his skateboard have a mass of 65.0 kg How much work is done by friction as the skateboard goes down the ramp?

What I used was
W=F(force) x r (displacement)
F= mv^2/r
= (65 kg x 9 m/s^2)/(5 m)
= 1035 N

W= (1035 N)(19.6 m) (To find 19.6, I used the circumference equation to find the distance down the ramp)

W= 20,286 J
20.3 kJ

I really do not think this is correct, and I'm trying to figure out what it means by "frictional" force.

Thanks for your help!
 
Last edited:
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Jim and his skateboard have gravitational potential energy at the top of the ramp, what is it? How much kinetic energy would he have if all of his potential energy were transformed into kinetic? How much kinetic energy does Jim have?

Remember
pe = mgh
ke = (mv^2)/2
 
so i don't have to use the radius of the circular (1/4) ramp at all??
 
The radius is also the height from which Jim starts.
 
oh my goodness- I'm so utterly confused... It's just not clicking :(!
 
If it were a straight ramp 5m high, would it be easier to understand?
 
yes, i think so- but that wouldn't be the same as it being curved, because the displacement on a curved ramp would be different than that on a straight ramp, no?
 
Not necessarily. If the ramp were a 45 degree incline, the displacement would be the same (5m down, 5m over); however, the way I figured this out does not need a displacement.

At the top of the ramp, Jim has no kinetic energy, but he has mgh potential energy. What does he have at the bottom?
 
mms05 is right, the displacement on a curved ramp would be different than that on a straight ramp, but, in this particular problem you can use a different approach , just like Teegvin said, in order to calculate work.

W = - (Uf - Ui)

where Uf is the final potential energy and Ui is initial potential energy.
 
  • #10
mms05 said:
Hi! I need some help with this question:

Jim rides his skateboard down a ramp that is in the shape of a quarter circle with a radius of 5.00 meters. At the bottom of the ramp, Jim is moving at 9.00 m/s. Jim and his skateboard have a mass of 65.0 kg How much work is done by friction as the skateboard goes down the ramp?

Find the total energy at the top and at the bottom. The difference in energy is the energy wasted to do work against friction when rolling down the ramp. :)
 
  • #11
thank you all! :)
 
  • #12
By conservation of total energy, initial total energy = final total energy + friction.
 

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