Calculating Thermal Power Produced by Friction for a Sliding Rock

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Hello, here is the problem I've been stuck on. It seems simple but I am getting the wrong answer. Any help would be appreciated.


A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200.

What average thermal power is produced as the rock stops?


My answers thus far are
784
1040
427

They were all wrong.
 
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Thesnail said:
Hello, here is the problem I've been stuck on. It seems simple but I am getting the wrong answer. Any help would be appreciated.


A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200.

What average thermal power is produced as the rock stops?


My answers thus far are
784
1040
427

They were all wrong.
You are right. They are all wrong. Besides, you don't have any units. What are the units of power here?

Perhaps you could explain your thinking first. How did you analyse the problem?

AM
 
Thesnail said:
Hello, here is the problem I've been stuck on. It seems simple but I am getting the wrong answer. Any help would be appreciated.


A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200.

What average thermal power is produced as the rock stops?


My answers thus far are
784
1040
427

They were all wrong.
How do you think this problem should be done? Your wrong answers do not tell us what you are thinking.
 
Thanks Andrew for responding.

It should be in W.

for DeltaX I did ((1/2)(.2)(9.8*20kg))/8ms that got me the distance 2.45m

Then I solved for acceleration and got 13.0612

Then I solved for force which was 13.0612*20kg = 261.224

I also solved for time which was .6125s

so I did V average of 2.45m/.6125s and got 4ms

So then I had everything I needed. to solve for Power average

P_av= F*V_av

So it was 261.224*4ms = 1044.9

But it said that was wrong? I don't know why.


Thanks, again for responding
 
Thanks you guys for responding to my question. I hope we can figure it out.
 
Okay, I solved it.

it was : ((mg)(friction)(x))/(time)

((9.80)(20)(.2)(2.45)/(.6125s) = 156.8 W
 
Thesnail said:
Okay, I solved it.

it was : ((mg)(friction)(x))/(time)

((9.80)(20)(.2)(2.45)/(.6125s) = 156.8 W

I am not sure how you determined x and time. One way to look at it is:

[tex]P_{avg} = Fv_{avg} = \mu_k mg\Delta v/2 = .2*20*9.8*4 = 156.8 W[/tex]

AM
 
i don't get it...
F=mg
V[tex]_{}ave[/tex]= v/2

but where did the [tex]\mu[/tex] came from?
 
Andrew Mason said:
I am not sure how you determined x and time. One way to look at it is:

[tex]P_{avg} = Fv_{avg} = \mu_k mg\Delta v/2 = .2*20*9.8*4 = 156.8 W[/tex]

AM

what equations did manipulate to arrive at this one [tex]P_{avg} = Fv_{avg} = \mu_k mg\Delta v/2 = .2*20*9.8*4 = 156.8 W[/tex]
 
F is not mg. F = umg or in other words the force of friction.