Net Work of Pulling 10kg Box Up 5m Incline

[dnt]

Homework Statement

a man pulls a 10 kg box up an incline at a constant speed to a vertical height of 5 m. what is the net work being done?

Homework Equations

w= fd

The Attempt at a Solution

i know in the plane perpendicular to the surface there is no net work done because all the forward forces (the pull) cancel out the backward forces (friction and gravity). however, isn't there a net force being done by gravity because he rose the box up 5 meters?

would the answer be (5m)(9.8)(10kg) = 490 J

or would it be 0 J because its not accelerating at all? whihc one is correct? thanks.

 

[Doc Al]

however, isn't there a net force being done by gravity because he rose the box up 5 meters?

No. Gravity is not the only force acting on the box. The net force is zero.

would the answer be (5m)(9.8)(10kg) = 490 J

That's the work done by the man, not the net work.

or would it be 0 J because its not accelerating at all?

Exactly.

 

[dnt]

thanks. what if the question did say he was accelerating up the slope with a = 5 m/s2? would that change the answer then?

 

[Doc Al]

thanks. what if the question did say he was accelerating up the slope with a = 5 m/s2? would that change the answer then?

Absolutely. In that case there would be net work done on the box.

 

[dnt]

would it be:

(10kg)(5 m/s2)(5m) = 250 J

is that right?

 

[Doc Al]

would it be:

(10kg)(5 m/s2)(5m) = 250 J

is that right?

That implies that the box accelerates straight up! The net work done depends on the angle of the incline and thus how far the force needs to act. For example, if the angle were 30 degrees you'd have to drag the box up 10 m to raise it 5 m vertically. So, the net work would be: (10 kg)(5 m/s^2)(10) = 500 J.

 

[dnt]

so basically you can take the net force (ma) in the same direction as its moving and then just calculate that distance to multiply to get the net work?

so gravity has no real part here?

 

[BishopUser]

Theres not enough information to solve the problem. Net work done on the object is equal to its change in kinetic energy

W_{net}=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2

to know the net work performed you need to know the initial velocity (did the box start from rest) and the angle of the ramp

 

[dnt]

but if i know the distance the box traveled i could also multiply that by ma (net force) to get work, right?

 

[BishopUser]

Make sure it is net distance (not just vertical), so if you are using W = F*D*Cos(theta) which is W = M*A*D*Cos(theta) those are all scalar values where theta is the angle between the force and distance vectors (distance vector should always be along the ramp). If the man is pulling along the ramp then theta = 0 and cos(theta) = 1. So if you are talking about distance along the ramp and the man is not pulling at an off angle then yes that should work

 

[Doc Al]

so basically you can take the net force (ma) in the same direction as its moving and then just calculate that distance to multiply to get the net work?

Yep.

so gravity has no real part here?

Gravity is already included in the net force.

but if i know the distance the box traveled i could also multiply that by ma (net force) to get work, right?

Sure. You don't need the initial velocity to find the net work done. (You would need the initial velocity to find the final kinetic energy.)