Silly work done on a car question.

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Homework Help Overview

The problem involves calculating the mass of a car given the work done to accelerate it from one speed to another. The context is rooted in the principles of work and kinetic energy in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between work and kinetic energy, with some attempting to apply the work-energy principle. Questions arise regarding the manipulation of equations and the use of velocity changes in calculations.

Discussion Status

There is an ongoing exploration of the correct application of kinetic energy formulas and the relationship to work done. Some participants express confusion about the equations and units, while others provide clarifications and guidance on how to approach the problem.

Contextual Notes

Participants note potential misunderstandings stemming from lecture notes and the need for clarity on the use of mass in the equations. There is acknowledgment of the complexity involved in transitioning from work to kinetic energy calculations.

MarcZZ
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Homework Statement



It takes 139 kJ of work to accelerate a car from 22.8 m/s to 27.7 m/s. What is the car's mass?

Homework Equations



Well, we know that work is also known as J and J is = to kg*m^2/s^2 so we can use the equation J = m(vf - vi) to solve this I think.

The Attempt at a Solution



So We know that 139 kJ is equal to 139x10^3 J. Also m(27.7 m/s - 22.8 m/s) = m(4.9m/s), so combing these we assume that 139000 J = ?kg(4.9m/s) But from here I'm getting confused badly. I think I'm misusing the equation somehow as I'm not sure how to manipulate the 4.9m/s in order to get rid of the kg*m^2/s^2 units on the other side.

Any help will be much appreciated. :)
 
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MarcZZ said:
Well, we know that work is also known as J and J is = to kg*m^2/s^2 so we can use the equation J = m(vf - vi) to solve this I think.
The work done will equal the change in the car's kinetic energy--not momentum. (That's why your units are not working out.) How would you calculate that?
 
KE = 1/2mv^2. I had pondered this, so in other words it's possible to use my change in velocity in place of V in the KE formula?
 
MarcZZ said:
KE = 1/2mv^2. I had pondered this, so in other words it's possible to use my change in velocity in place of V in the KE formula?
What you need is ΔKE = KEf - KEi.
 
I'm sorry but how can that be usable if you don't have the mass? Isn't it essentially equal to m = 2K / (delta V)^2. Sorry, I'm getting confused I just don't see this.

Edit: Never mind. I think I'm misunderstanding stuff because some notes I took in my lecture I miscopied. :/
 
MarcZZ said:
I'm sorry but how can that be usable if you don't have the mass?
You'll be using this to solve for the mass. (As you've probably figured out by now.)
 
Doc Al said:
You'll be using this to solve for the mass. (As you've probably figured out by now.)

Ahh never mind. I figure out what the problem was, I had used the formula you mentioned before but divided it by 2 as opposed to multiplying it. I've got it now the answer is 1.123x10^3 kg. Thanks very much for your help. :)
 

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