# Is this Power/Work problem Possible

#### littlepigboy5

I am currently in AP Physics C. I was given a homework problem that I don't think is possible with the given info.

A 20 kg rock slides on a rough horizontal surface at 8 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? (Answer: 157 WATTS) (first get the work done!)

Here I am given mass, velocity, force of friction( (20 * 9.8) * .2 = 39.2), AND THAT's IT. I tried getting work first, but I don't have a distance. I suppose I could solve for acceleration since friction is the only force in the X direction, which is -1.96, but even then, how would I get a distance from that. If I had a period of time this took place over I could find distance, but I don't. Could somebody explain this? Thanks

#### THERMO Spoken Here

Similar Problem...

Hi LPB,

Is it (v2)(v2) - (v1)(v1) = 2 ax? Does that get "x."

Interesting problem. I wonder if it is the average "frictional" power that happens?
Here's a friction loss problem. Not your problem but something to look at.

Crate Pushed Uphill | THERMO Spoken Here!

Good Luck, JP

#### littlepigboy5

Hi LPB,

Is it (v2)(v2) - (v1)(v1) = 2 ax? Does that get "x."

Interesting problem. I wonder if it is the average "frictional" power that happens?
Here's a friction loss problem. Not your problem but something to look at.

Crate Pushed Uphill | THERMO Spoken Here!

Good Luck, JP
I don't think that's it, but seeing that formula made me remember what to do. It's a kinematics equation. Vfinal (0) = vinitial(8) + acceleration(1.96)*time. From there I can solve for time and get the answer, thanks.

#### ChipB

PHF Helper
Right, the initial KE is all converted to heat, so the average power is

Power =$$\displaystyle \frac {\Delta KE} t = \frac {\frac 1 2 mv^2 }{t}$$

The time for the rock to stop comes from $$\displaystyle \Delta v = at$$. You just need to determine 'a' from $$\displaystyle F_r = \mu mg = ma$$.