How to Solve Energy & Work Problems?

  • Thread starter Thread starter dagitt
  • Start date Start date
  • Tags Tags
    Energy Work
Click For Summary
SUMMARY

This discussion focuses on solving energy and work problems related to a conveyor belt system and spring mechanics. The first problem involves calculating the power developed by a conveyor belt transporting 15 bundles of newspapers per minute, with each bundle weighing 28 kg. The second problem requires determining the work needed to stretch a spring from 0.14 m to 0.26 m using the area under a force vs. displacement graph. Key equations include W=F*d and P = change in W / change in T.

PREREQUISITES
  • Understanding of basic trigonometry
  • Familiarity with work-energy principles (W=F*d)
  • Knowledge of power calculations (P = change in W / change in T)
  • Ability to interpret force vs. displacement graphs
NEXT STEPS
  • Study the calculation of power in mechanical systems using the formula P = W/t
  • Learn about the work-energy theorem and its applications in physics
  • Explore how to analyze force vs. displacement graphs for work calculations
  • Review concepts of potential energy (PE = mgh) and kinetic energy (KE = 0.5mv^2) in problem-solving
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators seeking to clarify work and power calculations in practical scenarios.

dagitt
Messages
6
Reaction score
0

Homework Statement


1. A 10.5 m long conveyor belt, inclined at 30.0°, is used to transport bundles of newspapers from the mailroom up to the cargo bay to be loaded on to delivery trucks. Each newspaper has a mass of 1.0 kg, and there are 28 newspapers per bundle. Determine the power that the conveyor develops if it delivers 15 bundles per minute.

2. Use the graph in the diagram
2ilhhyx.jpg


to find the work needed to stretch the spring from 0.14 m to 0.26 m.

***

any or all help appreciated...THANKS SO MUCH <33


Homework Equations


For both 1 & 2...
basic trig, W=F*d, P = change in W / Change in T, KE = .5mv^2, PE = mgh (but I'm confused if I even need KE or PE to solve these...?)


The Attempt at a Solution


For 1, I drew out the ramp and all the force components. I determined the force per bundle to be 137.2 N per bundle, and the total Work to be 1440.6...I multiplied this by 15, to get the answer, but it wasn't correct...I'm just so confused as to how to approach this.

2. I added the change in force at those two points, and multiplied it by distance traveled which is .26 - .14 m. Got this wrong too...

any or all help appreciated. i suck at physics. thanks so much.
 
Physics news on Phys.org
For #1, are the units in your final answer correct? I assume they want it in Watts (J/s)

For #2, I'm not sure I understand what you are doing. What do you mean by "change in force at those two points"?
 

The Attempt at a Solution


For 1, I drew out the ramp and all the force components. I determined the force per bundle to be 137.2 N per bundle, and the total Work to be 1440.6...I multiplied this by 15, to get the answer, but it wasn't correct...I'm just so confused as to how to approach this.

This looks right to me.
2. I added the change in force at those two points, and multiplied it by distance traveled which is .26 - .14 m. Got this wrong too...

Work done is just the area under the graph, since W=Fd for constant F and d, and the graph can be broken down into an infinite number of infinitely thin rectangular tiles each with constant F and d.
 

Similar threads

Replies
10
Views
2K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
11
Views
3K
Replies
4
Views
3K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K