Help with thermal/kinetic friction energy?

[dorkymichelle]

Homework Statement

A child whose weight is 269 N slides down a 6.80 m playground slide that makes an angle of 41.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.190. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.314 m/s, what is her speed at the bottom?

Homework Equations

ke=1/2 mv²
thermal energy = f_kd = f_nU_kd

The Attempt at a Solution

for b I used 1/2mv_i² + mgh - f_kd = f_nU_kd = 1/2mv_f²
and got it correct...
I missed the class where the teacher explained thermal energy and frictional energy so I assumed that the energy lost to friction is turned into thermal energy and i used
thermal energy = f_kd = f_nU_kd
plugged everything in and got 228 J...
but maybe i have the relationship wrong?

 

[tiny-tim]

hi dorkymichelle!

(have a mu: µ )

I missed the class where the teacher explained thermal energy and frictional energy so I assumed that the energy lost to friction is turned into thermal energy …

yes, that's correct …

kinetic energy and potential energy are called "mechanical energy", and any "missing" energy (in questions like this) is thermal energy

 

[dorkymichelle]

so for a, I should just use equation thermal energy = fnUkd
and I got 228 J but when I put in the answer,(my homework is online) its wrong...
am i reading the question in correctly somehow?
but I got b right.

 

[tiny-tim]

hi dorkymichelle!

so for a, I should just use equation thermal energy = fnUkd
and I got 228 J but when I put in the answer,(my homework is online) its wrong...

i make it the same as you, 228 J (= 269*6.8*sin41°*0.19)

 

[dorkymichelle]

hi dorkymichelle!


i make it the same as you, 228 J (= 269*6.8*sin41°*0.19)

oops, found the mistake, in finding normal force, its mgCOS41 not Sin. =d silly mistake!