During a rockslide, a 520kg rock slides from rest down a hillside that is 500m long and 300m high. The coefficient of kinetic energy between the rock and the hill surface is .25.

Question

a) If the GPE energy U of the rock –Earth system is zero at the bottom of the hill, what is the value of U just before the slide?
b) How much energy is transferred to thermal energy during the slide?
c) What is the kinetic energy of the rock as it reaches the bottom of the hill?
d) What is its speed then?

a) I think I would use mgh =-1/2mv^2
b) How does one find the thermal energy?
c) I believe I would use KE=1/2mv^2
d) would I use the same equation as above?

a) U = initial potential energy = M g h
= (520 kg)*(9.8 m/s^2)*300 m = 1,529,000 J

b) Energy transferred to heat = (friction force) x (500 m)
= M g * 0.25 * cos A* 500 = 509,600 J
The hill angle A has a cosine of 0.8

c) subtract (b) from (a) to get KE

d) use KE = (1/2) m V^2 and solve for V

Answer 1

E(p)= E(pg) = mgh=1560000 J
E(k1)+ Epg(1)= E(k2) +E(pg2)
0+1560000=E(k2)+0 so E(k2)=1560000
Ek=0.5mv^2 so v= 77.46 m/s
W(f) = delta E(m) so
f*500*-1=E(m(f))-E(m(i))
so f = 1056.12N

Answer 2

heat released by thermal energy = 1560000J does not make sense. That is like saying that all of the energy was converted to heat!!!

Answer 3

heat released by thermal energy = 1560000J does not make sense. That is like saying that all of the energy was converted to heat!!!
And shouldn't we be using sin(A), not cos(A)??

Answer 4

Sorry, I just realized it is supposed to be cos(A)