Where Am I Going Wrong with My Water Slide Work and Energy Calculation?

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Homework Help Overview

The discussion revolves around a physics problem involving work and energy calculations for a student sliding down a water slide, where kinetic friction plays a role. The student is attempting to determine her speed at the bottom of the slide given her mass, the height of the slide, and the work done by friction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of the work-energy principle, questioning the signs and values used in the calculations. There are attempts to verify the correctness of the derived speed and the impact of significant figures on the final answer.

Discussion Status

The discussion is ongoing, with participants expressing uncertainty about their calculations and the correctness of their answers. Some suggest that the issue may relate to significant figures, while others are exploring the implications of the signs in their equations.

Contextual Notes

Participants note that the tolerance for the answer is +/- 2%, which raises questions about the precision of their calculations. There is also mention of the electronic homework submission system potentially influencing the perceived correctness of their answers.

bulbasaur88
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A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 70 kg, and the height of the water slide is 11.3 m. If the kinetic frictional force does -7.6 × 103 J of work, how fast is the student going at the bottom of the slide?

Could someone tell me where I am going wrong with this problem?

Vo = 0 m/s
m = 70 kg
Ho = 11.3 m
Hf = 0 m
Wnc = -7.6 x 103 J

Wnc = 0.5m(Vf2 -Vo2) + mg(Hf-Ho)
-7.6 x 103 = 0.5(70)(Vf2) + 70(9.8)(11.3)
Vf = 20.9 m/s
 
Last edited:
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I think that only your answer is wrong. All the rest is OK.
My answer is 2.08m/s
 
I tried your answer. That is incorrect also...Eeek please help!
 
That equation should have a positive sign in front of it...but still I get the answer wrong because my original answer is still 20.9 m/s
 
I wasted an attempt with 2.08 m/s. I have one try left please somebody help!
 
The tolerance allowed for this answer is +/- 2%. Does that mean I am just using the incorrect amount of significant figures ?
 
bulbasaur88 said:
Wnc = 0.5m(Vf2 -Vo2) + mg(Hf-Ho)
-7.6 x 103 = 0.5(70)(Vf2) + 70(9.8)(11.3)
Vf = 20.9 m/s
Hf - Ho should be negative.
 
With Hf - Ho = 0 - 11.3 = -11.3 meters, I get 2.08 m/s. This answer is still wrong according to the WileyPlus electronic homework submission site.
 
bulbasaur88 said:
With Hf - Ho = 0 - 11.3 = -11.3 meters, I get 2.08 m/s. This answer is still wrong according to the WileyPlus electronic homework submission site.
I'd say that answer is correct. As far as your online homework system goes, it may be looking for a different number of significant figures.
 

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