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

Click For Summary

Homework Help Overview

The problem involves a box being moved up a frictionless ramp by a constant force at an angle, with a specific focus on calculating the work done by that force over a vertical distance. The subject area encompasses concepts of work, forces, and motion on an incline.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the applied force and gravitational force, questioning the direction of work done. There is an exploration of the relevance of certain information, such as the angle of the incline and the implications of constant speed.

Discussion Status

The discussion is ongoing, with participants raising questions about the assumptions made in the problem setup and the calculations involved. Some guidance has been offered regarding the interpretation of forces and the relevance of specific details, but no consensus has been reached.

Contextual Notes

There is a noted lack of information regarding the angle of the incline, which may affect the analysis. Participants are also considering the implications of the box moving at constant speed in a frictionless environment.

edgarpokemon
Messages
93
Reaction score
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?
 

Attachments

  • mmm.gif
    mmm.gif
    2.5 KB · Views: 534
Physics news on Phys.org
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:
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.
 
Frictionless and the angle of the incline is not given !
 
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?
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
3K
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
18K
  • · Replies 1 ·
Replies
1
Views
12K