Calculating Heat Energy Generated from Work and Power

[husky88]

Homework Statement

A force of 120 N pushes a crate of mass 15 kg along the ground at a constant speed of 3.2 m/s for a distance of 6.2 m. What amount of heat energy was generated during this time?

Homework Equations

W=F*d
t=d/v
P=F*v
W=P/t

The Attempt at a Solution

I have solved this problem by finding out the work done by the force (W=F*d=744 J).
The time is t=d/v=1.9375 s.
The power that is actually used is P=F*v=384 W.
The work that is actually done is W=P/t=198.1935484 J.
So the heat energy generated is the difference between the two W amounts = 550 J.

I don't know if this is correct. I am also not using the mass given in the problem. I don't know if the mass is extra data, or if there is maybe a simpler way to do it. Maybe without using the Power, as there is no engine in the problem.
Any help or suggestions greatly appreciated. :)

 

[Mentz114]

A force of 120 N pushes a crate of mass 15 kg along the ground at a constant speed of 3.2 m/s for a distance of 6.2 m. What amount of heat energy was generated during this time?

The crate is being kept at constant velocity while the force is acting. What's stopping the crate from accelerating ? An opposing force which must be friction. So you need to calculate the frictional force, and this needs the mass of the crate.

 

[husky88]

How would I calculate the frictional force?
Ff=µ*Fn
I know Fn, but µ?
I was also thinking that since acceleration is 0, Fnet = 0.
So Ff = F = 120 N.
Therefore W = F * d. Just that simple.
Anyone agrees?

 

[Mentz114]

Yes. The opposing force must be equal to the applied force if the velocity is constant. So the frictional force is 120 N. Now you have it completely.

 

[husky88]

Thank you so much.
The problem sounds so complicated and confusing with a simple formula to use in the answer.

 

[Mentz114]

You reasoned your way to the correct answer, so you've had the most benefit you can get from the exercise. But the idea of friction dissipating energy as heat is also part of the answer. I hope you get full marks.