Calculating Work: Doubling and Halving Velocity of a 40 Kg Wagon

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Homework Help Overview

The discussion revolves around calculating the work required to change the velocity of a 40 kg wagon moving at a constant horizontal velocity of 10 m/s. Participants are exploring the relationship between work and kinetic energy in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the necessary information to solve the problem, particularly regarding the applied force and friction. There is also a discussion about the definition of work and its relation to kinetic energy.

Discussion Status

The discussion is ongoing, with some participants providing guidance on the definition of work and its application to the problem. There is an acknowledgment of the need for more information to proceed with calculations.

Contextual Notes

One participant notes the absence of specific values for applied force or friction, which are necessary to fully address the problem. The original poster expresses uncertainty about how to approach the calculations with the given information.

nikechristo
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This is the question :|

A 40 Kg wagon is moving with a constant horizontal velocity of 10 m/s.
1. How much work must be done to double this velocity?
2. How much work must be done to halve the original velocity?

SOmeone please help me. :|
 
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nikechristo said:
This is the question :|

A 40 Kg wagon is moving with a constant horizontal velocity of 10 m/s.
1. How much work must be done to double this velocity?
2. How much work must be done to halve the original velocity?

SOmeone please help me. :|

Welcome to the PF, nikechristo. We do not do your work for you here on the PF. You must show your work before we can offer tutorial help. That's why there is a Homework Help posting template (which you deleted) that you should use to post your questions. There is a spot at the end to show your work so far.

So show us how you will approach these problems. What is the definition of work, in terms of forces and other things?
 
Thanks for your welcome berkeman. :) Sorry for deleting the default template. I'm new to this forum so i didn't know the rules and regulations.

Work is done on an object whenever a force makes something move. This is the definition of work.

The general equation of work done is W = Force applied * distance.

Homework Statement

\

A 40 Kg wagon is moving with a constant horizontal velocity of 10 m/s.
1. How much work must be done to double this velocity?
2. How much work must be done to halve the original velocity?

Homework Equations



W = Force applied * distance


The Attempt at a Solution



I don't get this question. I need the applied force or the force of friction to solve this question, rite? I don't think there is enough information to solve this, but i may be wrong. Is there a way to find the work done by using the mass and distance?
 
The work done is the change in kinetic energy.
 

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