Calculate the work done moving a crate

Click For Summary
SUMMARY

The total work done in moving a crate involves calculating the work for each segment of the journey separately. The forces exerted were 34 N for 2 m, 13 N for 6 m, and -40 N for 2 m. The correct calculations yield W1 = 68 J, W2 = 78 J, and W3 = -80 J, resulting in a total work of 66 J, not the initially stated 189 J. This discrepancy arises from misunderstanding how to apply the work formula across different segments of motion.

PREREQUISITES
  • Understanding of the work-energy principle
  • Familiarity with vector addition of forces
  • Knowledge of basic physics equations, particularly Work = Force * Displacement
  • Concept of net force and its application in different segments of motion
NEXT STEPS
  • Study the work-energy theorem in detail
  • Learn about vector decomposition in physics
  • Explore examples of calculating work done in multi-segment motion
  • Review the effects of friction on work calculations
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone looking to deepen their understanding of work calculations in physics scenarios.

shepherd882
Messages
12
Reaction score
0

Homework Statement


you push a box out of a carpeted room and along a hallway with a waxed linoleum floor. While pushing the crate 2 m out of the room you exert a force of 34 N; while pushing it 6 m along the hallway you exert a force of 13 N. To slow it down you exert a force of 40 N through a distance of 2 m, opposite to the motion. How much work do you do in all?

Homework Equations


Work = Force * displacement

The Attempt at a Solution


Fnet = 34 + 13 + (-40) = 7 N
Displacement = 2 + 6 + 2 = 10m
Work = 7*10 = 70 J
(the answer's supposed to be 189 J; i thought about including friction for the carpet and linoleum floor for fnet, but there is no coeffiecient of friction given so I'm stuck)
 
Physics news on Phys.org
shepherd882 said:
Fnet = 34 + 13 + (-40) = 7 N
This is not correct. The force you are acting with varies with time. The forces you have added never act on the crate at the same time.
 
You can't determine an Fnet as the sum of the forces. You need to consider each segment of the journey separately and add the work from each to find the total.

With the scenario as stated, I don't see how they arrived at their answer of 189 J. Perhaps the problem was revised in some manner and the answer key not updated?
 
You must calculate the work separetly in this case (because the path took multiple directions). The box was puxed 2m into a direction by a net force in that direction of 34N:
W1 = 34 * 2
Then it was moved to another direction along 6m with a net force of 13N. W2 = 13 * 6
Lastly, it was applied a force of 40N contrary to the movement 2m. W3 = -40 * 2

The resulting work is the sum of all the works in the process: W1+W2+W3
 

Similar threads

  • · Replies 58 ·
2
Replies
58
Views
5K
  • · Replies 14 ·
Replies
14
Views
6K
Replies
1
Views
2K
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
29
Views
3K
  • · Replies 5 ·
Replies
5
Views
10K
Replies
25
Views
2K