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

In summary, the conversation discusses calculating the kinetic energy of a car with a mass of 1604 kg and traveling at a speed of 60 mph. The correct formula for calculating KE is provided and the question is raised about the total work done to reach this speed. The correct units for the calculation are discussed and a website is suggested to convert mph to m/s. The final conclusion is that the total work done is equal to the change in kinetic energy.
  • #1
connphysics
12
0
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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  • #2
What are your thoughts? Assume the car starts from rest. Note that your calculation for the car's KE is incorrect (units!).
 
  • #3
then wouldn't the total work be 2,887,200 J ? that IS the diff in KE I think !
 
  • #4
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.
 
  • #5
jay ! thanks so much ! could u help me convert them please !?
 
  • #6
For a start, 66 mph = 88 ft/s , exactly.
 
  • #7
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 ??
 
  • #8
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
 
  • #9
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.

[tex] E=\frac{1}{2} m v^2 [/tex]

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.
 

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

1. What is the Work-Energy Theorem?

The Work-Energy Theorem is a fundamental principle in physics that relates the work done on an object to its change in kinetic energy. It states that the net work done on an object is equal to the change in its kinetic energy.

2. How is the Work-Energy Theorem applied in real life?

The Work-Energy Theorem can be applied in various real-life situations, such as calculating the energy required to lift an object, determining the speed of a moving vehicle, or predicting the distance a ball will travel when kicked. It is also used in engineering and mechanics to design and optimize machines and structures.

3. Is the Work-Energy Theorem always true?

Yes, the Work-Energy Theorem is always true as long as there are no external forces acting on the object. This is because work and kinetic energy are both conserved quantities in a closed system.

4. Can the Work-Energy Theorem be used for non-conservative forces?

No, the Work-Energy Theorem only applies to conservative forces, which are forces that do not dissipate energy and follow the principle of conservation of energy. Non-conservative forces, such as friction or air resistance, do not follow this principle and cannot be accounted for using the Work-Energy Theorem.

5. What are the units of work and energy in the Work-Energy Theorem?

The units of work and energy in the Work-Energy Theorem are Joules (J) or Newton-meters (Nm). This is because both work and energy are measured in units of force multiplied by distance, and the standard unit of force is the Newton (N).

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