Solve Energy Conservation Problem: 17 kg Child Descending 3.5m Slide

[confusedbyphysics]

Please help me, this is due tonight at 11 and I can't figure out what to do!

" A 17 kg child descends a slide 3.5 m high and reaches the bottom with a speed of 2.5 m/s. How much thermal energy due to friction was generated in this process?"

The equation should be:

kinetic energy intiial + potential energy initial = ke (final) + pe (final) + friction X distance travelled

so

1/2mv^2 (initial) + mgy (initial) = 1/2mv^2 (final) + mgy(final) + Friction X d

The problem is I don't know d!

0 (no initial KE) + (9.8)(17)(3.5) = 1/2(17)(2.5)^2 + 0 (no final potential energy) + friction X d!

What is D!

I tried making the slide a right triangle with 45 degree angles for the other two and using the pythagorean theorem...and got 5 for the distance..but I got a wrong answer..

huh, could someone PLEASE help me!

 

[Seiya]

You don't need D to find out the answer, one second ill post my toughts.


ok let's see... forgive me if I am wrong...


Change E system = Change E mechanical + Change E thermal

E thermal is the friction and its equal to friction force times delta S

Change in mechanical energy is Eo-Ef... so 0 = Eo-Ef + E thermal

so

-E thermal = Eo - Ef
- E thermal = PEo - KEf
E thermal = Kef-PEo
E thermal = -530.025 J

Alright i hope this helps and anyone correct me if I am wrong... thanks

 

[confusedbyphysics]

Thanks. Hmm , my class hasn't learned how to do that yet. Is there any way to do it with the PE + KE = KE + PE + Fr (d)?

 

[Seiya]

thats exactly what I am doing except I am not using the distance...

(PEo + KEo) - (PEf+KEf) = - Work Friction (which is equal to Fr (d) ) Oo

if you need to do further work with the friction you can use the result i got and do

Work friction = Force friction (d)


:X

 

[confusedbyphysics]

Doh! You are right, seiya, the answer is 530. The friction work is Friction X d. For some reason I thought I was just looking for the force of friction. Stupid me.

Thank you for the help! :)

 

[Seiya]

you are welcome , just helping while thinking about my problem and hoping someone will give me a hint :p