Work Done by Gravity: Find Magnitude

Click For Summary
SUMMARY

The discussion focuses on calculating the work done by gravity on a 2.52 kg block pushed up a vertical wall. The user initially calculated the work using the formula W = mgh, resulting in -38.032J, which was deemed incorrect. The correct approach involves using the work formula W = Fd cos(theta), where the angle theta is 63.6 degrees between the applied force and the horizontal. The user is advised to reference their textbook for the proper terminology related to gravitational work.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of work in physics
  • Knowledge of force vectors and angles
  • Basic grasp of friction coefficients
NEXT STEPS
  • Review the concept of work-energy principle in physics
  • Study the calculation of work done by varying forces
  • Learn about vector components in force analysis
  • Explore the implications of friction on work calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of work calculations involving gravity and friction.

AdnamaLeigh
Messages
42
Reaction score
0
I'm sick of this stuff

A 2.52kg block is pushed 1.54m up a vertical wall with constant speed by a constant force of magnitude of F applied at 63.6 degrees with the horizontal. The coefficient of friction is .574. Find the magnitude of the work done by the force of gravity.

This is what I did:
W=mgh
W=2.52(-9.8)(1.52)=-38.032J

Apparently it's wrong and I don't understand why. I drew the diagram and the force of gravity is going down.

I already found F=38.648N and the work done by F, 53.265J for a previous problem but I don't think those results will help.
 
Physics news on Phys.org
Please tell me what mgh is equal to, and you will have solved your problem. (I don't mean the number, I mean look it up in your book and see what its called). :-)
 
AdnamaLeigh,
the work, W, done by a force, F, acting on a body while it is displaced through a distance d (a long a straight line), is given by
W = Fd cos(theta)
where theta is the angle between the the force and the displacement. The direction of the force and the displacement does not come into consideration in this calculation of the work.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
8K
Replies
10
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K