Work power and energy question


by mike_302
Tags: energy, power, work
mike_302
mike_302 is offline
#1
Jan14-08, 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
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
rock.freak667
rock.freak667 is offline
#2
Jan14-08, 05:07 PM
HW Helper
P: 6,214
All the work he is doing should be converted into the kinetic energy. So you can find v from the equation
lasershadow
lasershadow is offline
#3
Jan14-08, 05:31 PM
P: 15
Also, remember the work-kinetic 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.

mike_302
mike_302 is offline
#4
Jan14-08, 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.
belliott4488
belliott4488 is offline
#5
Jan14-08, 07:23 PM
belliott4488's Avatar
P: 662
Quote Quote by mike_302 View Post
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.
Heh ... thanks. I think the mentors on this site probably know what equations are relevant.

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?
belliott4488
belliott4488 is offline
#6
Jan14-08, 07:25 PM
belliott4488's Avatar
P: 662
Quote Quote by lasershadow View Post
Also, remember the work-kinetic 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.
Just to be clear: you don't have to do (a) and (b) first, since you know the amount of energy that is being converted to gravitational PE. You might as well answer them in order, though.
mike_302
mike_302 is offline
#7
Jan14-08, 07:34 PM
P: 103
Quote Quote by belliott4488 View Post
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.
Soooo, 2750J=Ek .... Ek = 1/2mv^2 .... Am I on a roll here?
mike_302
mike_302 is offline
#8
Jan14-08, 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?
lasershadow
lasershadow is offline
#9
Jan14-08, 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.
mgb_phys
mgb_phys is offline
#10
Jan14-08, 08:11 PM
Sci Advisor
HW Helper
P: 8,961
Correct, now for part c you need gravitational potential energy E = m g h
mike_302
mike_302 is offline
#11
Jan14-08, 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 ???
lasershadow
lasershadow is offline
#12
Jan14-08, 08:17 PM
P: 15
sadly, yes
mike_302
mike_302 is offline
#13
Jan14-08, 08:24 PM
P: 103
OK! Thanks you guys. c) Isn't to hard just because we had done something similar already.


Register to reply

Related Discussions
Physics : Work Energy Power Question Introductory Physics Homework 3
work/energy/power Introductory Physics Homework 1
Work, Energy and Power (Work Problem) Introductory Physics Homework 9
Work, energy and power question Introductory Physics Homework 4
PHYSICS work/power/energy question.. Introductory Physics Homework 3