How Is Work Calculated on a Box Moving Up a Frictionless Ramp?

In summary, a constant force F of magnitude 82.0 N is applied to a 3.00 kg shoe box at an angle of 53.0°, causing the box to move up a frictionless ramp at constant speed. To calculate the work done on the box by the applied force, the work equation cannot be used because the actual displacement of the box is unknown. Instead, the components of the force in the directions parallel and perpendicular to the ramp must be considered. The net force on the box is zero in the parallel direction, and the component of weight in the parallel direction is -mgsinθ.
  • #1
nothingsus
16
0

Homework Statement


A constant force F of magnitude 82.0 N is applied to a 3.00 kg shoe box at angle phi = 53.0°, causing the box to move up a frictionless ramp at constant speed. How much work is done on the box by when the box has moved through vertical distance h = 0.150 m?

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Homework Equations


W=Fdcos(theta)

The Attempt at a Solution


So I understand that you can use net work = W_g + W_f = change in KE = 0 (since constant speed) to solve this so W_f = mgd = 3*9.8*0.150 = , but I'm curious why you can't just use the work equation.

Here's my reasoning. Please explain where I'm incorrect.The applied force on the box acts at 53 degrees to the vertical and causes a displacement of 0.150m straight up. The angle between the force vector and the displacement vector is 180-53 = 127 degrees. so to calculate the work done on the box by the applied force only

W = 82*0.150*cos(127) = -7.40 J

It's negative (and also the wrong magnitude) which doesn't make sense because energy is supposed to be transferred to the box due to the force to move it upwards...?

Why can't I just use the work equation to find work done by a specific force only (the applied force in this case) like I have in other questions?
 
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  • #2
well i'd say that its because de box didn't move straight up.
It'd makes sense that if the box were moving straight up the work done by the force would be negative, since its opposing the box movement.

You can't use the work equation because you don't know the actual displacement of the box, the triangle with height 0.15 can have an infinite number of sides attached to it

Furthermore, you do the cosine of the angle, to find the force's component in the direction that the box moved. looking at the picture, the y component of the force in the y direction is downward (thats what you calculated~when doing Fcos(127)), the box doesn't move up due to the y component of the force, it moves up due to the x component.

Edit: to use the work formula here's what you'd have to do:
Find the component of F along the inclined surface, then find the component of that force with the vertical, and that should work...
 
  • #3
nothingsus said:
causing the box to move up a frictionless ramp at constant speed
What does this tell you about the net force on the box?
Edit: I see you have used W=mgh and I think my question is of no use. I believe since the angle of incline is unknown, you can't determine the angle between force and displacement.
 
  • #4
Let ##\theta## be the angle that the ramp makes with the horizontal. In terms of ##\phi## and ##\theta##, what are the components of the force F in the directions parallel and perpendicular to the ramp? Considering that the net force on the box is zero in the parallel direction, and the component of weight in the parallel direction is ##-mg\sin \theta##, what is the force balance on the box in the parallel direction?
 

Related to How Is Work Calculated on a Box Moving Up a Frictionless Ramp?

1. What is work done by an applied force?

Work done by an applied force is a measure of the amount of energy transferred when a force is applied to an object and causes it to move. It is a scalar quantity and is measured in joules (J).

2. How is work calculated?

The formula for calculating work done by an applied force is W = F * d * cosθ, where W is work, F is the magnitude of the applied force, d is the displacement of the object, and θ is the angle between the force and the displacement.

3. Can work be negative?

Yes, work can be negative. This occurs when the force applied to an object is in the opposite direction of the displacement, causing the object to lose energy. For example, when a person lowers a weight, the force applied by gravity is in the opposite direction of the displacement, resulting in negative work.

4. What is the unit of work?

The unit of work is the joule (J). However, it can also be expressed in other units such as newton-meter (Nm) or kilogram-meter squared per second squared (kg•m2/s2).

5. How does work relate to power?

Work and power are related because power is the rate at which work is done. Power is calculated by dividing work by time, or P = W/t. This means that an object that does a given amount of work in a shorter amount of time has a higher power.

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