What is the Total Work Done to Reach a Speed of 60 mph?

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Homework Help Overview

The discussion revolves around calculating the total work done to accelerate a car with a mass of 1604 kg to a speed of 60 mph, using the work-energy theorem. Participants are examining the kinetic energy calculation and the necessary unit conversions for speed.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of kinetic energy and question the correctness of unit conversions from mph to m/s. There is an exploration of the relationship between kinetic energy and total work done.

Discussion Status

Several participants have raised concerns about unit conversions and the accuracy of the kinetic energy formula. There is a mix of attempts to clarify the conversion process and the implications of incorrect calculations. Some express confusion about the topic's complexity and their placement in the forum.

Contextual Notes

Participants are working under the assumption that the car starts from rest, and there is an emphasis on ensuring proper unit conversions to apply the work-energy theorem correctly. There is also a mention of the original poster's frustration with the topic's difficulty.

connphysics
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if a cars mass is 1604 kg and it is traveling at a speed of 60 mph ...then it's KE is .5 X 1604 X 60 squared = a KE of 2,88,200 J ...now...my question is...using the work energy theorem - what is the TOTAL WORK done to get to a speed of 60 mph ??
 
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What are your thoughts? Assume the car starts from rest. Note that your calculation for the car's KE is incorrect (units!).
 
then wouldn't the total work be 2,887,200 J ? that IS the diff in KE I think !
 
Yes, the change in kinetic energy is the total work done. But your math is off (in the units), you forgot to convert mi/hr to m/s.
 
jay ! thanks so much ! could u help me convert them please !?
 
For a start, 66 mph = 88 ft/s , exactly.
 
can anyone help me understand this - I don't know how to convert 60 mph to ms

the original problem is - if a cars mass is 1604 kg and it is traveling at a speed of 60 mph ...then it's KE is .5 X 1604 X 60 squared = a KE of 2,88,200 J ...now...my question is...using the work energy theorem - what is the TOTAL work done to get to a speed of 60 mph ??
 
1 mile = 1.609 km = 1609 meters, 1 hour = 3600 seconds. so 60*1609 divided by 3600 gives you m/s which equals 26.81667 m/s
 
thanks all- but I'm more confused than ever now ! the original problem was - - if a cars mass is 1604 kg and it is traveling at a speed of 60 mph ...then it's KE is .5 X 1604 X 60 squared = a KE of 2,88,200 J ...now...my question is...using the work energy theorem - what is the TOTAL work done to get to a speed of 60 mph ??

can anyone here solve this ? maybe this is too advanced ? and i am posting it in the wrong topic ? i chose beginners physics cause that's the name of my class here in h school ! err- so frustrating !
 
  • #10
The problem is that the kinetic energy is worked out wrong.

E=\frac{1}{2} m v^2

In order for this to be in Joules the units need to be right. v cannot be in mph. It has to be in m/s, just as mass has to be in kg. All you need to do is convert mph to m/s and use the equation for kinetic energy.
 
  • #11
thank you so much ! finally making sense now ! but how do i convert mph to m/s ? is there a website that can do that for me ? thanks sooooo much
 
  • #12
You should be able to do it yourself. Currently you have miles per hour and you want metres per second. How many metres do you travel in one second if you travel a certain amount of miles in one hour?

If you travel 1 mile in an hour how many metres is that per hour, how many metres is it per second?
 
Last edited:
  • #13
Groundhog Day. Bueller?
 
  • #14
total work done = net change in KE.
 

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