Potential/conservation of energy

[kappcity06]

i got part of of this problem but b, c, and d I can't get. For c i think i am supposed to use k=1/2mv^2 but i don't know for sure? please help.

During a rockslide, a 340 kg rock slides from rest down a hillside that is 500 m long and 300 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.28.
(a) If the gravitational potential energy U of the rock-Earth system is set to zero at the bottom of the hill, what is the value of U just before the slide?
999600J
(b) How much energy is transferred to thermal energy during the slide?
? J
(c) What is the kinetic energy of the rock as it reaches the bottom of the hill?
? J
(d) What is its speed then?
? m/s

 

[Ukitake Jyuushirou]

helllo

is the ans for (b) 466301 J ?

 

[Doc Al]

Two (equivalent) ways to approach this:
(1) via energy: How much work is done against friction?
(2) via dynamics: What's the acceleration of the rock?

 

[sdekivit]

remember that no energy can be lost !

The amount of energy at the beginning of the movement equals that after the movement. Thus from the moment at rest till tdown at the hill:

$$E_{pot} = E_{kinetic} + E_{friction}$$
=
$$mgh = \frac {1} {2} \ mv^{2} + E_{friction}$$

 

[kappcity06]

thecanswer for b was not correct

 

[kappcity06]

i don't have a velicity to work with

 

[sdekivit]

i don't have a velicity to work with

that's what you need to calculate from this energy balance ;) For B $$E_{friction}$$ is asked.

$$E_{friction} = F_{friction} * s$$

 

[Office_Shredder]

(a) If the gravitational potential energy U of the rock-Earth system is set to zero at the bottom of the hill, what is the value of U just before the slide?
999600J
(b) How much energy is transferred to thermal energy during the slide?
? J
(c) What is the kinetic energy of the rock as it reaches the bottom of the hill?
? J
(d) What is its speed then?
? m/s

For b: What exactly is creating the thermal energy? I'll give you a hint... it starts with 'frictio' and ends with 'n'

Knowing this, is there any good way to calculate the total thermal energy created by the friction?

c: Once you've gotten b, what types of energy is the potential energy being converted to? Then just keep in mind conservation of energy, and you should be able to get the answer

d: Given the kinetic energy, velocity is easy to calculate

 

[Ukitake Jyuushirou]

thecanswer for b was not correct

not even close?

 

[kappcity06]

i don't know the answer to b yet

 

[kappcity06]

would i just use the coefficient of kinetic friction times the normal force for the energy of friction

 

[sdekivit]

would i just use the coefficient of kinetic friction times the normal force for the energy of friction

yes, but you need the work done by this force as you need to know the energy of friction that is lost as heat.

Hint: $$F_{N}$$ is not $$F_{g}$$

 

[HallsofIvy]

i got part of of this problem but b, c, and d I can't get. For c i think i am supposed to use k=1/2mv^2 but i don't know for sure? please help.

During a rockslide, a 340 kg rock slides from rest down a hillside that is 500 m long and 300 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.28.
(a) If the gravitational potential energy U of the rock-Earth system is set to zero at the bottom of the hill, what is the value of U just before the slide?
999600J

Yes. potential energy = mgh= (340)(300)(9.8) approximately.

(b) How much energy is transferred to thermal energy during the slide?
? J

A complicated calculation. You know the coefficient of friction and you can calculate the component of weight perpendicular to the hill (better draw some triangles to analyze that), then multiply by the coefficient of friction, 0.28, to get the actual friction force. Multiply the friction force by the length of the slope (use the Pythagorean theorem to get that) to find the energy expended in sliding down the slope. Of course, that work becomes thermal energy (heat) in the hillside.

Added later: No! It's not at all complicated. The energy used up in friction (which becomes thermal energy) is just the coefficient of friction times the Potential energy! That's remarkable. I had to do the calculation for any slope before I believed it.

(c) What is the kinetic energy of the rock as it reaches the bottom of the hill?
? J

You know the potential energy at the top of the hill and, since the rock was sitting still, it had no kinetic energy so that (your answer to (a)) is the total energy. Since the rock has potential energy 0 at the bottom of the hill, part of the total energy has gone into thermal energy (part (c)) and what is left is kinetic energy.

(d) What is its speed then?
? m/s

Kinetic energy is (1/2)m v². You are given m and you know the kinetic energy from (c). Solve for v.

 

[kappcity06]

oookk. I got it now. thanks to all who helped

 

[kappcity06]

i didn't get it

 

[sdekivit]

i didn't get it

lol, what you didn't get

 

[kappcity06]

for b i did 340 kg* 9.8 (for the force perpendicular to the hill) times the coefficient of kinetic friction, .28 then I muiltiplyed that by the lengght of the hill.583.1

i got 544008.976 could this be right?

 

[sdekivit]

for b i did 340 kg times the coefficient of kinetic friction, .28 then I muiltiplyed that by the lengght of the hill.583.1

we have a hill here with a certain angle. The vector of the normal force stand in a right angle at the surface.

--> you need to make 2 vectors of the gravitational force: 1 in the direetion of the movement and 1 at a right angle at the surface.

The angle can be calculate by the length of the hill and the highth use goniometry.

Thus: $$F_{n} = F_{g} * cos \alpha$$

 

[kappcity06]

so one vector directed down the hill at 9.8 and one vector directed perpendicular to the hil at 3332

 

[kappcity06]

i got an angle of 59 degress tan-1(500/300). then i did 9.8cos59=5.04

 

[sdekivit]

so one vector directed down the hill at 9.8 and one vector directed perpendicular to the hil at 3332

yes. The gravitational force is easy to calculate. Then you need to know the angle of the hill and this also the angle of one of the 2 new vectors you made out of the vector of the gravitational force.

Thus use the equation i posted before about the normal force.

--> i get for the angle:

$$sin \alpha = \frac{300} {500}$$

$$\alpha = sin^{-1} (0,6) = 36,9$$

 

[kappcity06]

so the normal force is 1713.6 N

 

[kappcity06]

yes. The gravitational force is easy to calculate. Then you need to know the angle of the hill and this also the angle of one of the 2 new vectors you made out of the vector of the gravitational force.

Thus use the equation i posted before about the normal force.

--> i get for the angle:

$$sin \alpha = \frac{300} {500}$$

$$\alpha = sin^{-1} (0,6) = 36,9$$

I'm not sure what this means

 

[sdekivit]

so the normal force is 1713.6 N

Sorry. I don't get that:

$$F_{N} = 340 * 9.81 * cos 36.9 = 2667.3 N$$

 

[kappcity06]

ooooo 36.9

 

[kappcity06]

ok so know 2667.3 times .28=746.844
that isthe energy from friction. is that the answer

 

[sdekivit]

ok so know 2667.3 times .28=746.844
that isthe energy from friction. is that the answer

please forgive me: I'm from holland and we use the , instead of the . in these numbers. So if you see any ,s you now know they should be a .

Now that's force of friction and we need energy of friction. Go back to a few replies earlier.

 

[kappcity06]

so 746.844 times s will give us the energy of friciton. I don't know what s stands for.

dont worry about the comma thing

 

[kappcity06]

is s the side length 583.1

 

[sdekivit]

so 746.844 times s will give us the energy of friciton. I don't know what s stands for.

dont worry about the comma thing

s = distance of the movement, thus the length of the hill: 500 m

 

[kappcity06]

no its 500 m long and 300 m high

 

[kappcity06]

using a^2+b^2=c^2 i found it to be 583.1

or should i use the 500

 

[kappcity06]

ok that right sry

 

[kappcity06]

ok part b is now correct for part c i need to find kentic enery.

 

[sdekivit]

using a^2+b^2=c^2 i found it to be 583.1

or should i use the 500

you mean that the horizontal length is 500 m ? then $$F_{N}$$ must be recalculated with angle = $$tan^{-1} \frac{3} {5}$$

the rest is the same.

I thought the length of the hill you meant the hypotenuse of the triangle