How Does a Student's Work Affect Motion on a Scooter?

[mike_302]

Homework Statement

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?

Homework Equations

Ek=1/2mv^2 ?

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

 

[rock.freak667]

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

 

[lasershadow]

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]

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]

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]

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]

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]

ORRR, wait a tick, another equation I have is W=delta E so... E=Ek

W=Ekmax - Ekmin(zero)
W=Ekmax
W=1/2mv^2Sub it all in and we get approx. 9.00m/s for v ... Does that sound correcT?

 

[lasershadow]

Right

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

 

[mgb_phys]

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

 

[mike_302]

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

 

[lasershadow]

sadly, yes

 

[mike_302]

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