Kinetic energy - this girl is getting frustrated

[physicsgirl8]

this is a 100 level college course ! it isn't supposed be this hard !

anywayyyy - here is the question

need some help- problem - the cars mass is 1604 kg- calculate its KE at 60 miles per hour...so- I converted mph to meters per second and got 60 mph = 26.82 meters per second - so - the KE formula is .5 times 1604 times 26.82 squared = 576,888.54

So the KE would be 576888.54 correct ? and what would the unit be for this ? Joules ??


anyone help me on this ?

 

[Pengwuino]

Yes, it's correct and it's in Joules.

And by the way, remember no one here knows what university you go to so saying it's a 100 course doesn't mean anything to anyone :)

 

[physicsgirl8]

thank you sooooo much pengwuino! yay ! my hero ! lol

one more for you ...the next question is "apply the work -energy theorem to find the TOTAL WORK DONE to get to a speed of 60 miles per hour"

could you puhleeeeese help me with this one ?

 

[physicsgirl8]

or anyone else for that matter! Thx all ;)

 

[lewando]

Well, what do you know about the work energy theorem? Where are you getting stuck?

 

[terberculosis]

Joules are a unit of work. Assuming the car starts from rest, the kinetic energy of the car is the amount of work needed to accelerate the car to that velocity.

Work = force * distance (kg*m*m)/(s*s) and Kinetic Energy = .5 m v v (kg m m)/(s s )

 

[physicsgirl8]

all I know is that according to the work-energy theorem, if one or more external forces act upon a rigid object, causing its kinetic energy to change from Ek1 to Ek2, then the mechanical work (W) done by the net force is equal to the change in kinetic energy. For translational motion, the theorem can be specified as:[3]

W = \Delta E_k = E_{k_2} - E_{k_1} = \tfrac12 m (v_2^2 - v_1^2) \,\!

where m is the mass of the object and v is the object's velocity.

If a force F that is constant with respect to time acts on an object while the object is displaced in a straight line along the length and direction of a vector d, the mechanical work done by the force on the object is the dot product of the vectors F and d:[4]

W = \bold{F} \cdot \bold{d} = F d \cos\theta

 

[physicsgirl8]

this stuff is tough ! They are asking me to find the total work done to get to a speed of 60 mph...wouldn't that be the same as the KE ...in other words - 576,888.54 J ? I'm confuzed ;)

 

[terberculosis]

Yes it would be the same as the KE. Assuming the car started from rest and the force is in the direction of displacement.

 

[physicsgirl8]

thanks terb! one more ?? the next question is "Calculate the average power during the acceleration period of 9.9 seconds"

anyone that can give me some tips on this final part of the assignment ?!

u guys r great !

 

[terberculosis]

Power is work divided by time.

 

[physicsgirl8]

terb...sooooo...576,888.54 divided by 9.9 seconds = 58,271.57 ...but what is the SI unit ...watts ? so 58 kilowatts would be the answer approx ?

 

[terberculosis]

sounds good to me.

 

[physicsgirl8]

nice...so- final final question...and this is just for me...this car (it's called the nissan leaf) it seems by our calculations tonight - that it's power is approx 58 kilowatts...but nissan advertises it's engine power at 80 kilowatts...can anyone think of a reason why there could be a difference between the two values ?
terbs ? ;)

 

[terberculosis]

OK. The engine may be able to put out that much power, but does all of it go to increasing the car's velocity? How well does the rest of the car get the engine's energy to the ground and turned into velocity?

 

[physicsgirl8]

you're a genius ;) thanks for helpin ;)