What is the final speed of the box at 10m?

Click For Summary

Homework Help Overview

The problem involves a force exerted by a rope lifting a 2.0 kg mass vertically over a distance of 10 m, with a force that varies from 20 N to 30 N. The original poster seeks to determine the final speed of the box at this height.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses confusion regarding how to approach the problem due to the non-constant force and acceleration. Some participants suggest using work-energy principles and conservation of energy, while others clarify the concept of work as the area under the force-distance graph.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and offering guidance on relevant concepts such as work and energy. There is no explicit consensus yet on the approach to take.

Contextual Notes

There is a mention of a lack of calculus in the course, which may limit the methods available for solving the problem. The original poster also notes a specific answer, which may influence the direction of the discussion.

avsj
Messages
26
Reaction score
0

Homework Statement



A force is exerted by a rope lifting a 2.0kg mass a vertical distance of 10m from the ground. A graphs shows F (N) vs d (m). THe F starts at 20 N and goes to 30 N in a straight line as d goes from 0 to 10m. What is the final speed of the box at 10m?

Homework Equations



F=ma
kinematics equations

The Attempt at a Solution



I am thoroughly lost as to how to deal with this one. The force applied is not constant, so the acceleration will also not be constant. I'm guessing there is another way of looking at it... would really appreciate a hint in which direction to start... thanks.. the answer is 7.3m/s by the way
 
Physics news on Phys.org
Work is equal to the integral over distance of force. Use conservation of energy to compare this with mgh and then use KE=mv^2/2.
 
what do you mean work is equal to the integral over distance of force, what is the integral?
 
avsj said:
what do you mean work is equal to the integral over distance of force, what is the integral?

No calculus in this course? Then if you graph force versus distance, work is the area under the curve. That's what an integral is.
 

Similar threads

What is the Work Involved in Lifting and Moving a Box with a Crane?
  • · Replies 4 ·
Replies
4
Views
2K
Work Check - Centripetal force - Finding Tension in a Rope
  • · Replies 8 ·
Replies
8
Views
3K
Worker Sim Pulma Sheen: Solving a Pulley Problem
  • · Replies 6 ·
Replies
6
Views
2K
Force and Energy of a block on a spring as it compresses.
  • · Replies 3 ·
Replies
3
Views
2K
Lifting an object, kinetic energy, work and forces
  • · Replies 11 ·
Replies
11
Views
6K
Does the mass of an object affect its inertia?
  • · Replies 6 ·
Replies
6
Views
3K
How to find speed with Faraday's law
  • · Replies 7 ·
Replies
7
Views
2K
Newton's laws of motion: A gun firing bullets....
  • · Replies 16 ·
Replies
16
Views
3K
Trouble dealing with vector coordinates in question
  • · Replies 2 ·
Replies
2
Views
1K
Work Pulley Problem with constant speed
  • · Replies 2 ·
Replies
2
Views
1K