Work Needed to Accelerate 2008 Dodge Viper SRT

  • Thread starter Thread starter lolcatz
  • Start date Start date
  • Tags Tags
    Accelerate Work
AI Thread Summary
To determine the work needed for a 2008 Dodge Viper SRT to accelerate from rest to 30 m/s, the kinetic energy formula KE = 1/2 mv² is appropriate. With a mass of 1560 kg, the calculation yields the work done as KE = 1/2(1560 kg)(30 m/s)². There is some confusion regarding the distinction between work and power, as work is measured in joules, while power is measured in watts. Additionally, the discussion highlights the importance of considering time when calculating power, particularly in relation to part B of the question, which asks for power developed over 4 seconds. Overall, the key focus is on applying the correct formulas for work and power in the context of the Viper's acceleration.
lolcatz
Messages
3
Reaction score
0

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?
 
Physics news on Phys.org
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:

W = F \cdot \Delta d

P(t) = F(t) \cdot v(t)

P = \frac{W}{\Delta s}

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; W, or in other terms:

Js^{-1};

kg \cdot m^{2} s^{-3};

N \cdot ms^{-1}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 E_{k}
 
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 =]
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top