Work Needed to Accelerate 2008 Dodge Viper SRT

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Homework Help Overview

The discussion revolves around calculating the work needed to accelerate a 2008 Dodge Viper SRT from rest to a speed of 30 m/s, given its mass of 1560 kg. Participants are exploring concepts related to kinetic energy and work-energy principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the kinetic energy formula, questioning whether the resulting units are appropriate for work. There is also mention of different formulas related to power and energy transfer.

Discussion Status

Several participants are engaged in clarifying the relationship between work and energy, with some confirming the use of kinetic energy as a means to calculate work. There is an ongoing exploration of the implications of time on power calculations, particularly in relation to the initial question.

Contextual Notes

Participants note that this is a take-home quiz, and there are specific formulas provided for the assignment, including kinetic energy, potential energy, work, power, and efficiency. There is some confusion regarding the units of work and the relevance of time in the calculations.

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Homework Statement



The 2008 dodge viper srt uses a v-10 engine to develop more than 600HP of the power,

question- If the mass of the viper is 1560 kg, what work must be done by the car to accelerate from rest to 30m/s>


Homework Equations



I am confused on what formula to use


The Attempt at a Solution



I did.. KE=1/2mv^2... so...1/2(1560KG)(30m/s)^2... is this correct?
 
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lolcatz said:

Homework Statement



The 2008 dodge viper srt uses a v-10 engine to develop more than 600HP of the power,

question- If the mass of the viper is 1560 kg, what work must be done by the car to accelerate from rest to 30m/s>

Homework Equations



I am confused on what formula to use

The Attempt at a Solution



I did.. KE=1/2mv^2... so...1/2(1560KG)(30m/s)^2... is this correct?

Yes.

Out of interest, what other formulae do you have at your disposal?]

edit: I didn't read your question properly, I do apologise.

this is the energy transfer, not power.

For power there are several equations available:

[tex]W = F \cdot \Delta d[/tex]

[tex]P(t) = F(t) \cdot v(t)[/tex]

[tex]P = \frac{W}{\Delta s}[/tex]

and also remember that horsepower can be converted to SI units of power.
 
Last edited:
its a take home quiz... the formulas stated are... KE, PE, work, power and %lost

i was hesitant because Joules obtained from the KE formula are not units of work? am i correct?
 
lolcatz said:
its a take home quiz... the formulas stated are... KE, PE, work, power and %lost

i was hesitant because Joules obtained from the KE formula are not units of work? am i correct?

That is correct, the units of work are Watts; [tex]W[/tex], or in other terms:

[tex]Js^{-1}[/tex];

[tex]kg \cdot m^{2} s^{-3}[/tex];

[tex]N \cdot ms^{-1}[/tex]As you can see, all of these units involve a measurement of time, so are you sure there is not mention of time in the question?

edit: you are confusing me haha; the units of work ARE joules, so the work done is the [tex]E_{k}[/tex]
 
Last edited:
well ...


part B of the question says... if the car does this work in a time of 4 seconds, what power did the car develop?
could you use this on part A?
 
haha, read my edits please =]
 

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