# Power constant, box up ramp

• chrispy2468
In summary, the conversation discusses the calculation of work and power while pushing a 10 kg box up a rough ramp at a constant speed of 2 m/s. The rate of work done on the box by the person pushing is determined by the formula P=Fv, where F is the force and v is the velocity. The force of friction, gravity, and the net force are also taken into consideration in the calculations. The net force on the box is zero since it is moving at a constant velocity.

## Homework Statement

You push a 10 kg box up a rough ramp at a constant speed of 2 m/s.
uk=0.4
Theta=15
vi-vf=0 since constant which means a=0
Questions are
a. What is the rate at which you do work on the box? b. What is the rate at which gravity does work on the box? c. What is the rate at which friction does work on the box? d. What is the rate at which the net force does work on the box?

## Homework Equations

I am not certain believe I should find distance traveled first..
Then the rest of work might be simple..
W=F*d
Wyou=mg sin(0)(yf-yi)-Ff
Wgravity=mg cos 180(yf-yi)
ma+mgsin(theta)-ukmgcos(theta)
P = F*v = rate of work done
x

## The Attempt at a Solution

Ffric= uk*mgcos(theta)= .4(10kg)(9.8)cos(15)=.380N?

which I think would = 0 so that must be wrong
I think I am on the wrong track, any help would be appreciated. Thank you!

[/B]

You are correct that acceleration is zero ... so d=vt
You are not asked to find the work, you are asked to find the rate of work ... which is power, which you have given as P=Fv
You know v, so you just want F ... remember though that the direction of the force and the velocity counts too.

Thank you for the clarification!.. so..
Pyou=mgsin(0)-ukmgcos(15) *V
Pg=mgcos(90)*v
Pfric=ukmgcos(15)*V
Pforcetotal=Fyou-Fg-Ffric*V
??

Remember that movement in the opposite direction of the force makes the work negative.
The actual equations are ##W=\vec F\cdot\vec s## and ##P = \vec F\cdot\vec v## (where ##\vec s## is the displacement vector).

Note: since the box moves at a constant velocity, what is the total (net) force on the box?