How Is Work Calculated and Power Determined in Physics Problems?

[!!!!!!!]

Mmmm...Good ol' work and power

A 1200N force is applied parallel to a horizontal surface. It pushes an 80N box 10m across the surface. What work is done? What power is dissipated in 3 minutes?

What i did for this one is subtract the 80N force from the 1200N force and multiply it by 10m because W= FxD
1200-80=1120x10m=11200J, problem is I am not sure if I am supposed to subtract the 80N force of the box. For power, i did 11200/180s=62.22 watts, but I am not sure if that's just power or if its power dissipated
any help would be greatly appreciated, thanks

 

[BishopUser]

You should not be subtracting the 80N from 1200N. The 80N does have some significance, but in this situation the force vector of the 80N (its weight) is perpendicular to the displacement vector so it will come out to zero work. What I'm trying to say is that since it is not on a sloped plane, the work done by gravity is 0. Therefore, the only work done is by the 1200N force.

Also remember that the equation for work is $$w=\vec{F}\cdot\vec{D}$$ which means $$w=|F||D|cos\theta$$ theta being the angle between the force and the displacement vectors.

 

[!!!!!!!]

So ill just find power by multiplying 1200 by 10m to get 12000J, but what about the power dissipated? i found power to be 66.67 by dividing work by time, but I am not sure if that's the dissipation or not

You should not be subtracting the 80N from 1200N. The 80N does have some significance, but in this situation the force vector of the 80N (its weight) is perpendicular to the displacement vector so it will come out to zero work. What I'm trying to say is that since it is not on a sloped plane, the work done by gravity is 0. Therefore, the only work done is by the 1200N force.

Also remember that the equation for work is $$w=\vec{F}\cdot\vec{D}$$ which means $$w=|F||D|cos\theta$$ theta being the angle between the force and the displacement vectors.

 

[BishopUser]

You get the work by multiplying the two. I don't think the word "dissipated" has any significant meaning other than how much power was exerted/used/etc, I believe all they are asking for is the standard work/time for power.

 

[!!!!!!!]

Cool, thanks a lot man :)

You get the work by multiplying the two. I don't think the word "dissipated" has any significant meaning other than how much power was exerted/used/etc, I believe all they are asking for is the standard work/time for power.