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Work power and energy question 
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#1
Jan1408, 04:59 PM

P: 103

1. The problem statement, all variables and given/known data
A Student, starting from rest, does 2750 J of work to propel himself on a scooter across level ground. The combined mass of the scooter and the student is 68.0 kg. Assuming no friction, find... a) How fast is he travelling? b) What is his kinetic energy? c) If he then coasts up a hill, to what height does he rise before stopping? 2. Relevant equations Ek=1/2mv^2 ??? 3. The attempt at a solution Not even sure where to start on this given only 3 completely different variables. We can figure out c) if we are given some sort of start on a) and possibly b). Thanks in advance 


#2
Jan1408, 05:07 PM

HW Helper
P: 6,202

All the work he is doing should be converted into the kinetic energy. So you can find v from the equation



#3
Jan1408, 05:31 PM

P: 15

Also, remember the workkinetic energy theorem. Work is change in kinetic energy. For C, you can use conservation of energy with Kinetic all being transferred to gravitational potential. But, first you need to do a and b.



#4
Jan1408, 07:15 PM

P: 103

Work power and energy question
Just note (we're trying to do this equation as you reply) that the "relevant equation" I gave is likely not hte only one that we haev to use. Others such as Eg=mgh, Mt=Mtprime, and so on, are likely to be involved.



#5
Jan1408, 07:23 PM

P: 666

Just think about the energy relations in this problem. The work that the student does is initially (i.e. before he reaches the hill) entirely converted to KE, so you can just equate the two. After he's rolled up the hill to a stop, all the KE (which is equal to the work he did at the start) is now converted to gravitational PE. You have the equations giving these quantities in terms of things you know. Can you identify the relevant equations now? 


#6
Jan1408, 07:25 PM

P: 666




#7
Jan1408, 07:34 PM

P: 103




#8
Jan1408, 07:54 PM

P: 103

ORRR, wait a tick, another equation I have is W=delta E so... E=Ek
W=Ekmax  Ekmin(zero) W=Ekmax W=1/2mv^2 Sub it all in and we get approx. 9.00m/s for v .... Does that sound correcT? 


#9
Jan1408, 08:09 PM

P: 15

That's right.
Now, parts b and c follow logically. All of the student's kinetic energy becomes gravitational potential. 


#10
Jan1408, 08:11 PM

Sci Advisor
HW Helper
P: 8,953

Correct, now for part c you need gravitational potential energy E = m g h



#11
Jan1408, 08:17 PM

P: 103

So, just to confirm then (because it sounds far too easy) Ek (the asnwer for b)) will be 2750 J ???



#12
Jan1408, 08:17 PM

P: 15

sadly, yes



#13
Jan1408, 08:24 PM

P: 103

OK! Thanks you guys. c) Isn't to hard just because we had done something similar already.



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