Why Is Work the Same for Different Distances on an Inclined Plane?

Click For Summary
SUMMARY

The discussion centers on the concept of work in physics, specifically regarding inclined planes. It establishes that the work done (W = fd) remains constant regardless of the distance traveled, with a specific example showing that increasing the distance from 0.5 to 4 reduces the necessary force by a factor of 8. The minimum force required to lift an object vertically (F1 = mg) is compared to the force required to move it up a ramp (F2 = mgsinθ), leading to the conclusion that F2 equals F1 divided by 8. This relationship clarifies the misunderstanding regarding the horizontal component of force when analyzing work on an inclined plane.

PREREQUISITES
  • Understanding of basic physics principles, particularly work and force.
  • Familiarity with inclined planes and their mechanics.
  • Knowledge of trigonometric functions, specifically sine.
  • Basic algebra for manipulating equations.
NEXT STEPS
  • Study the concept of work-energy principle in physics.
  • Learn about the mechanics of inclined planes in detail.
  • Explore the relationship between force, distance, and work in various scenarios.
  • Investigate the application of trigonometric functions in physics problems.
USEFUL FOR

Students preparing for the MCAT, physics enthusiasts, and educators looking to clarify concepts related to work and inclined planes.

lovelyrwwr
Messages
48
Reaction score
0
Screenshot_2013-10-05-04-37-13.png
W = fd
The answer is 8. The solution manual says work is the same for either scenario, d = 4 or d = 0.5

From this assumption I can see that by increasing distance from 0.5 to 4, you reduce the force necessary by a factor of 8.

However what I cannot comprehend is: where does the assumption that work is the same for both scenarios come from? Since you have the added horizontal component to consider when pushing it up the ramp, isn't that extra work? Or...Have I labeled the triangle incorrectly? Should the 4 be labeled on the other leg rather than the hypotenuse?Sorry if this is a silly question but I haven't taken intro physics in a couple years and am currently studying for the mcat. It seems not all of the principles of physics have hstayed with me lol. Thank u in advance pf!

Homework Equations


The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
You can analyze this problem in this way

The min force required to lift an object of mass m vertically up from bottom to top of the step is F1= mg

The min force required to move an object of mass m up along the ramp from bottom to top of the step is F2= mgsinθ

From this you can deduce F2=F1/8 .

Does this help ?
 
Yasssss! Thank you so much!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
3
Views
2K
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K