What is the work rate of an escalator lifting a woman from bottom to top?

  • Thread starter Thread starter nelly42688
  • Start date Start date
  • Tags Tags
    Work
Click For Summary
SUMMARY

The work rate of an escalator lifting a 50.0 kg woman from the bottom to the top can be calculated using the formula W = mg(hf - hi). Given that the escalator is 30.0 meters long and moves at a speed of 1.00 m/s, the height (h) can be derived from the length and angle of the escalator. The work done by the escalator is directly related to the gravitational force acting on the woman, which is 490 N (50.0 kg * 9.81 m/s²). The power output can then be determined by dividing the work done by the time taken to ascend.

PREREQUISITES
  • Understanding of basic physics concepts such as work and power
  • Familiarity with gravitational force calculations
  • Knowledge of kinematic equations related to motion
  • Ability to interpret and manipulate formulas in physics
NEXT STEPS
  • Calculate the height of the escalator using trigonometric functions based on the angle and length
  • Learn about the relationship between work, energy, and power in physics
  • Explore the concept of mechanical advantage in inclined planes
  • Study the effects of friction and efficiency in escalator systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of work and power calculations in real-world applications.

nelly42688
Messages
10
Reaction score
0

Homework Statement



An escalator is 30.0 meters long and slants at relative to the horizontal. If it moves at 1.00 m/s, at what rate does it do work in lifting a 50.0-kg woman from the bottom to the top of the escalator

im not to sure on how to go about doing this problem, the question doesn't make logical sense since its an escalator without an angle... anyone have insight on this or am i just not seeing it.
 
Physics news on Phys.org
i no i didnt add anything except for the problem but any tips or help will push me off my feet because there's something i just don't see here

i geuss you would use the equation .5mv^2 +mgh=.5mv^2 + mgh which tehre is no H in this case so it would bsaically be a standard
W=mg(hf-hi) which in this case would just be mg?
 
Last edited:
anyone know? i don't want you to do the problem for me just i need help getting started, I looked through my physics book and all my notes and i can't find a formula that would satisfy this.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 13 ·
Replies
13
Views
10K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K