# Work and Energy of water slide

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

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I think that only your answer is wrong. All the rest is OK.

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 ?

Doc Al
Mentor
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.

Doc Al
Mentor
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.