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

[nchin]

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!

 

[frogjg2003]

You didn't factor in the change in potential energy.

 

[nchin]

You didn't factor in the change in potential energy.

im not sure what you mean. do u mind showing me the steps?

 

[frogjg2003]

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.

 

[Nickie]

For part b, you can use the conservation of energy equation again, but this time use the speed at the top of the slide (0.534 m/s) as the initial speed and the final speed at the bottom as the unknown.

So, we have:

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

108.410 J = 1/2m(v final)^(2) - 0.5(267)(0.534)^(2)

Solving for v final, we get:

v final = 5.45 m/s

Therefore, the child's speed at the bottom of the slide is 5.45 m/s.