Work done using time, velocity and mass?

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Homework Help Overview

The discussion revolves around calculating the work done on a vehicle as it decelerates from 90 km/h to 50 km/h over a period of 8 seconds, applying concepts from the law of conservation of energy.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of the work-energy principle and constant acceleration equations. There are attempts to clarify the correct application of formulas and the significance of initial and final velocities. Questions arise regarding the assumptions made about deceleration and the relevance of certain equations.

Discussion Status

The discussion includes various attempts to substitute values into the work formula, with some participants questioning the correctness of the initial calculations. There is acknowledgment of a potential misunderstanding regarding units and conversions, and some participants express confusion about the required approach based on the context of the problem.

Contextual Notes

Participants note that the problem originates from a paper that does not require the use of constant acceleration equations, leading to discussions about the appropriateness of different methods. There is also mention of a specific solution provided in the paper, which adds to the complexity of the discussion.

TDizzl
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Homework Statement



The law of conservation of energy applies to the motion of vehicles. Find the work done on a 1200kg vehicle when it slows from 90kmh-1 to 50kmh-1 in 8.0 seconds.

Homework Equations



W=1/2mv^2 - 1/2mu^2

The Attempt at a Solution


[/B]
W=0.5*1200*50^2 - 0.5*1200*50^2
 
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If you assume a constant deceleration, you can use constant acceleration equations such as x = 1/2(vo + v)*t and v = vo + a*t. Work = Fd, where F is the force and d is the displacement. This can be expanded to W = m*a*d. Using the acceleration equations, you should be able to find all of your unknowns and plug them into the work equation.
 
TDizzl said:
W=0.5*1200*50^2 - 0.5*1200*50^2
Is that what you meant to post? What happened to the 90kmph?
cnstntcnfsn said:
If you assume a constant deceleration...
Why would TDizzl care what the rate of deceleration is?
 
Yes, my bad
Its W=0.5*1200*50^2 - 0.5*1200*90^2
 
cnstntcnfsn said:
Using the acceleration equations

The question is from a paper which does not require the use of constant acceleration equations. It would seem bizarre to use a formula that hasn't been introduced to us yet.
Also, the solution is written as -2.59*10^5 Joules in the paper.
 
TDizzl said:
Yes, my bad
Its W=0.5*1200*50^2 - 0.5*1200*90^2

You're not proposing to leave that as your answer, are you?
 
PeroK said:
You're not proposing to leave that as your answer, are you?
No, that was how I substituted the values in as an attempt to answer the question.
 
TDizzl said:
Yes, my bad
Its W=0.5*1200*50^2 - 0.5*1200*90^2
In what units?
 
haruspex said:
In what units?
m=1200kg
v=50kmh-1
u=90kmh-1
 
  • #10
TDizzl said:
m=1200kg
v=50kmh-1
u=90kmh-1
So what is it in Joules?
 
  • #11
haruspex said:
So what is it in Joules?
Yea its like -3360000J, that's what I got
 
  • #12
TDizzl said:
Yea its like -3360000J, that's what I got
No, the speeds you are given are in kmh-1, not ms-1.
 
  • #13
So how do I get the answer as -2.59×105?
 
  • #14
TDizzl said:
So how do I get the answer as -2.59×105?
How do you convert km/h to m/s?
 
  • #15
Alright -2.59×105 is the answer I got, the problem was that I was doing 502/3.6 instead of (50/3.6)2.
Thanks for the help, feeling extremely stupid. :mad:
 

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