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.
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
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
heat released by thermal energy = 1560000J does not make sense. That is like saying that all of the energy was converted to heat!!!
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)??
Sorry, I just realized it is supposed to be cos(A)
a) To find the value of U just before the slide, you can use the equation for gravitational potential energy (GPE) given by U = mgh, where m is the mass of the rock, g is the acceleration due to gravity, and h is the height of the hill. The GPE energy U of the rock-Earth system is given as zero at the bottom of the hill. So, you can calculate U by multiplying the mass of the rock (520 kg) by the acceleration due to gravity (9.8 m/s^2) and the height of the hill (300 m):
U = (520 kg) * (9.8 m/s^2) * (300 m) = 1,529,000 J
b) To find the amount of energy transferred to thermal energy during the slide, you need to consider the work done by friction. The work done by friction is given by the equation work = force * distance. In this case, the force of friction can be calculated using the equation force = coefficient of kinetic friction * normal force. The normal force is equal to the weight of the rock, which can be calculated as the mass of the rock (520 kg) multiplied by the acceleration due to gravity (9.8 m/s^2). The normal force has a cosine component of the hill angle A. Given that the coefficient of kinetic friction is 0.25 and the distance is 500 m, you can calculate the energy transferred to heat as follows:
Energy transferred to heat = force * distance
= (coefficient of kinetic friction) * (normal force) * (distance)
= (0.25) * (mass of the rock) * (acceleration due to gravity) * (cos A) * (distance)
= (0.25) * (520 kg) * (9.8 m/s^2) * (cos A) * (500 m)
Let's calculate cos A using the equation cos A = adjacent side / hypotenuse. The adjacent side is given as 300 m and the hypotenuse is given as √(300^2 + 500^2). Hence, cos A = 0.8.
Energy transferred to heat = (0.25) * (520 kg) * (9.8 m/s^2) * (0.8) * (500 m)
= 509,600 J
c) The kinetic energy of the rock as it reaches the bottom of the hill can be found using the equation for kinetic energy (KE), which is given by KE = 1/2mv^2, where m is the mass of the rock and v is its velocity. To find KE, we need to subtract the energy transferred to heat (calculated in part b) from the initial potential energy U (calculated in part a):
KE = U - energy transferred to heat
= 1,529,000 J - 509,600 J
= 1,019,400 J
d) To find the speed of the rock at the bottom of the hill, you can use the equation for kinetic energy (KE) mentioned in part c. Rearranging the equation, we have:
KE = 1/2mv^2
v^2 = (2KE) / m
v = √((2KE) / m)
Plugging in the values of KE (1,019,400 J) and m (520 kg), we can calculate the speed v:
v = √((2 * 1,019,400 J) / 520 kg)