Conservation of energy problem child slides down at angle? check my asnwers?

AI Thread Summary
A child weighing 267 N slides down a 6.10 m slide at a 20° angle, with a coefficient of kinetic friction of 0.10. The initial calculations for thermal energy transfer were incorrect, as the correct formula is μN multiplied by the slide length, resulting in 153 J. For the final speed at the bottom, the change in potential energy must be considered, as it affects the kinetic energy. The correct approach involves accounting for both the work done against friction and the loss of potential energy. The final answers are 153 J for thermal energy and 5.45 m/s for the child's speed at the bottom.
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A child whose weight is 267 N slides down a 6.10 m playground slide that makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.10.

(a) How much energy is transferred to thermal energy?

(b) If she starts at the top with a speed of 0.534 m/s, what is her speed at the bottom?

my attempt:

a)267sin30° - 0.1 x 267 cos 20° = 108.410 J

b) ƩW = 1/2m(v final)^(2) - 1/2m(v initial)^(2)

108.410 J = 1/2m(v final)^(2) - 2.845J

111.255 = 1/2m(v final)^(2)

...v final = 2.858 m/s

The answers are
a) 153 J
b) 5.45 m/s

im not sure what I am doing wrong. can someone please help?

thanks!

EDIT:

I figure out what i did wrong for part a.

the answer is just mu n
0.1x267cos(20)x6.1 = 153.047

I still don't understand part b!? help please!
 
Last edited:
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You didn't factor in the change in potential energy.
 
frogjg2003 said:
You didn't factor in the change in potential energy.

im not sure what you mean. do u mind showing me the steps?
 
The child slides down the slide. That means her height decreased by some amount. She's under the influence of gravity, so she lost some potential energy. The work by friction becomes W=ΔKE+ΔPE.
 
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