How Much Work Does Friction Do on a Crate Sliding Down a Ramp?

  • Thread starter Thread starter cbeeson23
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on calculating the work done by friction on a 70 kg crate sliding down a 4.0-meter ramp inclined at 30 degrees. The crate reaches a speed of 5.0 m/s at the bottom, leading to the conclusion that the work done by friction is -500 Joules. The calculation involves the equation Mgh - Wf = 1/2MV^2, where the gravitational potential energy is converted into kinetic energy, and friction opposes this motion, resulting in a negative work value for friction.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of work and energy
  • Knowledge of gravitational potential energy calculations
  • Basic proficiency in algebra for solving equations
NEXT STEPS
  • Study the principles of energy conservation in physics
  • Learn about the work-energy theorem and its applications
  • Explore the effects of friction on motion in different scenarios
  • Investigate the relationship between potential energy and kinetic energy
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of motion and energy dissipation due to friction.

cbeeson23
Messages
7
Reaction score
0
A 70 kg crate starting at rest slides down a rough ramp 4.0 meters long, inclined at 30 degrees above horizontal. At the bottom, it is moving at 5.0 m/s. what was the work done by friction on the crate?

Answers=
A. -500 Joules
B. 500 Joules
C. 460 Joules
D. -460 Joules
E. 0.0 Joules

I got this answer wrong and i was wondering why... so here is my work. i got the height to be 2meters by using sin 30

Mgh - Wf = 1/2MV^2

70 * 9.81 * 2 - wf = .5 * 70 * 5.0^2
875 - wf = 1373.4
wf =1373.4-875
wf = 500 joules

so i got positive 500 joules. So is the answer actually -500 joules since friction forces go up the ramp>
 
Physics news on Phys.org
The question is worded poorly, IMO. The proper phrase should be "how much energy is dissipated by friction?" The ambiguity is in the definition of 'work'. Since the frictional force acts to oppose the motion, the frictional force can be thought of as removing some fraction of the total gravitational potential energy available to perform work.

Of course, the energy is removed as heat and plastic deformation of the materials.
 
cbeeson23 said:
Mgh - Wf = 1/2MV^2
Instead, think:
Total Work done by all forces = ΔKE
Wg + Wf = ΔKE
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
1
Views
2K
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
10K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
5K