Work Done by Weight Homework: Find W

AI Thread Summary
The discussion revolves around calculating the work done by weight on a 700 kg crate on an inclined surface. Initially, there was confusion about the term "work done by weight," which refers to the work done by gravity, represented as the change in potential energy (ΔPE). After clarifying this, the focus shifted to finding the work done by friction when additional parameters were introduced, such as a new final velocity and time. Participants emphasized the need to apply conservation of energy principles to solve for the frictional force. The conversation highlights the importance of understanding the concepts of work, energy, and forces in physics problems.
Legerity
Messages
19
Reaction score
0

Homework Statement



A 700 kg crate is on a rough surface inclined at 30°. A constant external force P = 5600 N is applied horizontally to the crate. The force pushes the crate a distance of 3.0 m up the incline, in a time interval of 7.3 s, and the velocity changes from v1 = 1.4 m/s to v2 = 2.5 m/s. What is the work done by weight?

75512f21-629f-400f-914b-f9f928f9ba49.png


Homework Equations



W = Fd, KE = 0.5mv^2, PE = mgh

The Attempt at a Solution



I drew the free-body diagram, but I am stumped on how to go about the problem. What is the work done by weight, and how can I find it? I don't exactly know what it is.
 
Last edited:
Physics news on Phys.org
I don't know what they mean by 'work done by weight'. My guess is they meant the work done by the force P. (Cut and paste error from another question?)
 
haruspex said:
I don't know what they mean by 'work done by weight'. My guess is they meant the work done by the force P. (Cut and paste error from another question?)
I figured out the problem, and it is the work done by gravity on the box (ΔPE).
 
Legerity said:
I figured out the problem, and it is the work done by gravity on the box (ΔPE).
Really? So most of the information is irrelevant? (Or maybe there are more parts to the question?)
 
haruspex said:
Really? So most of the information is irrelevant? (Or maybe there are more parts to the question?)
Yes, there is another part of the question I am stuck on. It is now asking me to find the work done by the frictional force, and my attempts at the problem have been futile. The only things changed in the info in the original post is that now V2 = 2.3 m/s, and time = 8.3 s. Could you possibly help?
 
Last edited:
You'll have to use an equation that relates conservation of energy and the energy lost by friction: KE_{i}+PE_{i}+WE_{i}=KE_{f}+PE_{f}+WE_{f}
 
Legerity said:
It is now asking me to find the work done by the frictional force,
You know the distance, so it remains to calculate the frictional force. What equations do you get from your free body diagram?
 
Legerity said:
I figured out the problem, and it is the work done by gravity on the box ( - [/color]ΔPE).
see important correction in red[/color].
 
Back
Top