[Physics] What is the force when you are pushing at a constant velocity?

Question

The whole question is:
"An 80.0kg box is pushed at a constant velocity along a frictionless incline as shown in the diagram. (sorry no diagram, I'm gonna describe it) How much work is done on the box in moving it from the bottom to the top of the incline?

Diagram: it would be easier if you drew this out from that im saying.
right angle triangle
7.0m height
10.0m hypotenuse

I know that you use W=Fd(cos x), but how do I figure out F when they say constant velocity?

Answer 1

force (F) exerted by agency in moving the box is important. Also, what is the state velocity/acceleration of box while applying that F
let us write the equation of motion of box along the plane

m a = net up force = F (applied upwards) - weight along plane downwards
m dv/dt = F - [m* g sin x]
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mg weight splits into (mg sin x) along plane and mg cos x normal to plane which provides reaction for stability
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given v = constant so dv=0=change in velocity
0 = F - m* g sin x
F = applied = m* g sin x
work done = F * displacement along force vector
W = m* g sin x * 10 meter
W = 80*9.8* [7/10]*10
W = 5488 Joule
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had this box been moved with some acceleration up the plane then our F would have been different (more force needed to impart acceleration)
ma + mg sin x = F(with acceleration) so agency would have done more work

Answer 2

Work done by a force is found by the product of the force and the distance along which the force acts.

In the problem the weight of the box is 80 x9.8 = 78.4 N The hypotenuse is 10 m and height is 7m.

If we were not provided with an inclined plane, we will lift the box by applying a vertical force of 78.4 N and will move this force through a distance of 7m.

The work that the force will do is 78.4 x 7 = 548J
Having provided with an inclined plane we can do the same work by applying a less force but moving it through a longer distance.

Since the hypotenuse is 10 m, the force applied need to be displaced through 10m... Instead of 78.4 N, now it is suffice to apply a force of 54.8N and move the box through 10m. The work done is 54.8 x 10 = 548 J. This is the same as the previous one.

The inclined plane is a simple machine which helps us to convert the force and its direction as per our convenience. But the total work that we have to do is the same and we cannot change it.

To find how much the force can be changed, depends upon the geometry of the plane.

The hypotenuse and height are connected by the simple relation sinθ = h/ L where L is the hypotenuse.

The work done = mg * h. but h = L sin θ
The work done = mg L sin θ.

This suggests that instead of moving the force mg through a distance of h, we can as well apply a force of mg sin θ and move it through a distance of L. In both cases the work done remains the same.

When the inclined plane is horizontal, the force mg acts vertically down, when we increase the angle of the plane, the normal to the plane also inclines from the vertical direction through this angle.

A component of the weight mg sin θ now acts along the inclined plane and tends to pull it down along the inclined plane. To make the object move up the plane we have to apply a force of mg sin θ along the plane. (Note that in the case of vertical lift we apply a force of mg only)

Answer 3

In the case of constant velocity, the total force applied in the horizontal direction is 0. Any force applied in the horizontal direction to push the block up the incline is canceled by the equal and opposite reaction force from the incline.

Therefore, the only work done will be in the vertical direction.

80kg * 9.8 m/s^2 * 7 m = 5488 N*m

Answer 4

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Answer 5

the force you are excerting is to over come gravity. so there is a vertical force = 80kg x 9.81m/s^2 due to the weight of the box.
You need to figure the portion of that force you ar e overcoming, which is related to the angle of the plane.
Or you could just set the work equal to the change in potential energy...