Work problem, box moves at h=.25 with F=85N

In summary: The box moves up a frictionless ramp at a constant speed. Fa is the force that is applied to the box and it causes it to move. The box moves up a frictionless ramp at a constant speed and the box uses Fa to move. The box moves up a frictionless ramp at a constant speed and the box uses Fa to move.
  • #1
edgarpokemon
93
1

Homework Statement



A constant force Fa of magnitude 85.0N is applied to a 3.0kg shoe box at angle 55.0, causing the box to move up a frictionless ramp at constant speed. How much work is done on the box by Fa when the box has moved through vertical distance h=.25m?

Homework Equations


mgsin(angle)-Fa=0(constant speed)
mgsin(55)=Fa=24.08N
hyp=opp/sin(55)=.305m
Work=+7.35J (problably minus because the work is pointing down and displacement is up)
mgh=-7.35J

The Attempt at a Solution


So I know that the simple way is just to use mgh. But in my homework the the box is going up, so I assume that the gravity will do negative work. But the answer is positive and they used mgh. They used Wa=-Wg and they they switched the -Wg to +Wg, and i am not sure why they did that. So was what I did correct too?
 

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  • #2
Hello Ed,

I don't see ##m\vec g\sin\theta## and ##\vec F_a## pointing in the same direction ? I agree that the component of ##\vec F_a## along the incline should be 24.1 Newton (my ##|\vec g| = 9.81## m/s2). And if ##\vec F_a## does no work perpendicular to the incline (there is no displacement in that direction), but only along the incline, over the given distance, I get a different result !

[edit] I suspect this exercise is a lot more involved then you (and I at first) think. Where in the drawing is the ##55 \ ^\circ## ? Is ##\phi = 55^\circ##. In that case: check: "frictionless" and "at constant speed" !
[edit2] suspicion relieved, see below o:) . And I don't get a different answer any more . But ##Fa## is an unknown.
 
Last edited:
  • #3
BvU said:
[edit] I suspect this exercise is a lot more involved then you (and I at first) think.
I would suggest it is rather less involved than it appears. The trick is to realize what information is irrelevant.
Ed, think of it from the perspective of the box and its overall change of state.
 
  • #4
Frictionless and the angle of the incline is not given !
 
  • #5
BvU said:
Frictionless and the angle of the incline is not given !
Is that in response to my post? If so, what point are you making?
 

Related to Work problem, box moves at h=.25 with F=85N

1. How do you calculate work in this scenario?

To calculate work, you would use the equation W = F*d, where W is work, F is force, and d is distance. In this case, the force is 85N and the distance is 0.25 meters, so the work would be 21.25 joules.

2. What is the unit of measurement for work?

The unit of measurement for work is joules (J).

3. How does the height of the box affect the work done?

The height of the box does not directly affect the work done. However, it may affect the amount of force needed to move the box and the distance the box is moved, which would then impact the work done.

4. What is the relationship between work and energy?

Work and energy are closely related. Work is the transfer of energy from one object to another, and energy is the ability to do work. In this scenario, the work done on the box results in an increase in the box's kinetic energy.

5. How does the force applied affect the work done?

The force applied directly affects the work done. The greater the force applied, the more work is done on the box. In this scenario, a force of 85N is applied, resulting in 21.25 joules of work being done on the box.

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