Mechanics (work) questions

and finally
A man of mass of 70kg rides a bike of mass 15 kg at a steady speed of 4m/s up a road which rises 1m for every 20m of its length. What power is the cyclist developing if there is a constant resistance to motion of 20N?

I calculated the change per second in GPE and the work done against friction but do i need to include kinetic energy of the biker as well as the GPE, because this mucks up my results

The first two don't need to be worked out, just answer the queries but i'd be thankful if someone could work out the 3rd one

Power = Work / Time

or more acurately P = dW/dt

Now the work done on an object equals its change in energy. So:

W = ΔEk + ΔEp + Losses due to friction

Now technicaly one should have a negative sign there, but what we are really trying to calculate is not the work done on the biker, but rather the work done by the biker and thsu the two minuses cancel.

Now, the questions states that the biker is traveling at constant speed. Thus the change in velocity Δv = 0. Now this gives:

ΔEk = 1/2m(u)² - 1/2m(v)² = 1/2m ( u² - v² )

Now since Δv = 0 , u = v and so ( u² - v² ) = 0
Hence
ΔEk = 0

So tehre is no change in kinetic energy and the work reduces to:

W = ΔEp + Losses due to friction

hence the average power output is just:

P = (ΔEp + Losses due to friction)/ Δt

Or if you use calculus:

P = dw/dt = d(Ep + Losses due to friction)/dt

This is true no matter what units you use, but if you want the power in Watt , use Joules for energy and seconds for time... Then:

P Watts = d/dt (Ep + Losses due to friction) Nm/s

ah thanks
so i could say P = (85g *1 + 20)/5 = 170.77W

i did (85g - 20)/5 initally but then i started wondering about kinetic energy and extra downwards forces et al

and to the other guy - thanks as well

1.) A car of mass 800kg moving at 20m/s is brought to rest by a braking force of 800N

a) calculate the work done by the brakes

2.) A pump (power 600W) raises water through a height of 3m and delivers it with a velocity of 6m/s.
What mass is removed in 1 minute?

3.) A man of mass of 70kg rides a bike of mass 15 kg at a steady speed of 4m/s up a road which rises 1m for every 20m of its length. What power is the cyclist developing if there is a constant resistance to motion of 20N?

1.) m = 800kg, v = 20, F = -800 N.

Power (Brakes) = |F| * v = 800 * 20 = 16, 000 W.

F = ma
---> -800 = 800a
---> a = -1.0 ms^-2.

u = 20, v = 0, a = -1.0:
v = u + at
---> 0 = 20 - t
---> t = 20 s

Work Done (Brakes) = Power * Time = 16, 000 * 20 = 320, 000 J.

2.) P = 600 W, h = 3 m, v = 6 ms^-1, t = 1 min = 60 s.

Work Done = Power * Time = 600 * 60 = 36, 000 J.

Work Done = Gain In GPE.

---> 36, 000 = mgh
---> 36, 000 = m(9.8)(3)
---> Mass Removed = 36, 000/(9.8 * 3) = 1224.5 kg

I amde a mistake with the 2nd part I think, sorry.

yeah, but your first part kicks mines ass

right, I reconsidered that 2nd bit, yes, theres the KE to consider as well: the work done is used to undergo the height against gravity, plus giving the water its KE.

So that part is done.

let me have a look

1.) but using your method E = 320,000J and then s = 400m (W=Fs)

2.) for the first one i just worked out the KE to be 160,000J and said it would be conserved

1.) yes, thats assuming that 320, 000 J is used to fully stop the car - that certainly may well not be the case (and can't be assumed) during braking as heat energy is probably lost due to friction against the road.

2.) again, you can't assume conservation during braking because energy is most certainly lost as heat energy due to the friction of the tyres against the road.