# Homework Help: Work, energy, power

1. Jul 11, 2012

### roovid

work, energy, power helpp

question:

A man of mass 70kg rides a bicycle of mass 15kg at a steady speed of 4ms-1 up a road which rises 1m for every 20m of its length. what power is the cyclistdeveloping if there is a constant resistance to motion of 20N?

Ans: 250W

My sol'n so far:

Sin∅ = 1/20
in 1second object moves 4m
therefore
1/20 = x/4
x = 0.2m

Ep = mgh
=85(10)(0.2)
=170

Ek=0.5mv2
= 0.5(85)42
= 680

Ek+Ep= 170 + 680
= 850

work done by resistive force = fs
= 20(4)
= 80

Wcyclist= [ Ep + Ek ] + Wresistive force
= 680 + 80

stuck
plz help
where am i going wrong

Last edited: Jul 11, 2012
2. Jul 11, 2012

### TSny

Note, the general concept is Wcyclist + Wresistance = ΔEk + ΔEp.

Note those Δ's!. They mean "change in". The speed of the cyclist is "steady". So what is the change in kinetic energy as the cyclist goes up the hill?

Also, is the work done by the resistance force positive or negative? (Hint: What is the direction of the resistance force relative to the direction of motion?)

Once you get the work, how do you get the power?

3. Jul 11, 2012

### roovid

how do i find del E now?
i do not understand how to proceed from what u said
resistive force will be -ve
plz help

4. Jul 11, 2012

### Simon Bridge

You can see your mistake by checking the units for each of your calculations ...

Try another approach:

Work = force times distance, therefore:
Power = force times speed.

5. Jul 11, 2012

### roovid

THANK U
i think i got it out

used the component of weight parallel the to the displacement axis for the F grav and it worked out.
thnx again

6. Jul 11, 2012

### roovid

w parallel to plane
85(10)(1/20)
=42.5

p = (42.5 + 20)4
= 250W

7. Jul 11, 2012

### TSny

Great! Simon's suggestion was a very good one (although I don't see anything wrong with your units - maybe I'm just not seeing it.)

The same answer follows from the work-energy approach:

Wcyclist + Wresistance= ΔEk + ΔEp

Wcyclist - 80 J = 0 + 170 J.

So, Wcyclist = 250 J. Since this is the work done in 1 second, the power is 250 J/s or 250 W.

But the "F times v" method gets the answer with less effort.

8. Jul 11, 2012

### Simon Bridge

I really wanted to point out that 170+80=250 but that would have been the same as handing out the answer :)

Note on units: I had interpreted them:
Ep=mgh but what he calculated was mgv ... units J/s
Eres=20Nx4m/s (i.e. Fv)... units: J/s
EK=0.5mv2 ... units J
... spot the odd one out :)
...of course, I realise - these were just energy calculations for 1 second in time ... and he did say, once, but I missed the import. I don't like to encourage the "work it out it for one second" thing because that can obscure some of the concept - like here where it would imply that moving costs you 0.5mv2 every second!

It is usually more constructive to adopt a "work out the equation first then bung in the numbers" approach that I had, hopefully, modeled.

9. Jul 11, 2012

### TSny

Simon, OK. Good.

10. Jul 11, 2012

### roovid

oh lol
thnx for the input
much appreciated

my prob was i wasnt noticing that speed was constant, so the would have no change in the kinetic value

Last edited: Jul 11, 2012
11. Jul 12, 2012

### Simon Bridge

<mutter>mind you I also prefer energy arguments to force ones</mutter>
Ho well it's all good ... cheers.