Recent content by physiman

Thanks for the hint.For Ice -> water: m = 0.1kg c=2100 J kg^-1 K^-1 dt = 10 degrees (-10 -> 0) therefore E= mc dt = 0.1(2100)(10) = 2100 J Energy required to melt the ice: E= specific latent heat of fusion of ice * mass = 325000(0.1) = 32500 J and then for water to steam: m = 0.1 kg...

1. Homework Statement Calculate the heat energy required to convert 0.10 kg of ice at -10 degrees celsius into steam at 100 degrees celsius. 2. Homework Equations E = m c dt (1) where E = heat energy (kJ) m = unit mass (kg) c = specific...

Oh cool, I did get ~1523.41 N and from the 1223.41 worked out for friction it would support F=ma. Thanks a lot, you were a real help. If there was a reputation system I would show my appreciation :)

Ok, that must be the Horizontal force then, so 600(9.81)sin15 instead of using cosine?

oh alright, sorry. tbh apart from the friction I can't think of any force, things like air resistance should be negligible

Not mentioned on the Question but the only forces I can think of is Weight of the car (perpendicular to the track, therefore 600(9.81)cos15 ), the frictional force and there should be a Normal reaction too.

How can I get to the Friction force from working out the resultant force? Oh yh, and about the energy lost to friction bit: Do I just work out the Net energy from before motion and after the car has traveled 100m? i.e.: Initial KE: 0 J Initial GPE: mgh = 152340.8899...J Total Energy...

Did you use W=Fd? Because my answer (1523.4 N) doesn't look far away from your one.

I used mgh. But I worked out the height from the ground, because GPE depends on it. So I just used 100sin15 and got about 25.9 m for it.

9.81 to be precise ;)

That wasn't mentioned, it says it 'free-wheeled' down the hill. Would that mean the engine was off?

Yes, sorry forgot to mention that. The time given was 20s.

First, a big hello to the Physics Forums community :) Ok, now to business: A car with mass 600 kg accelerates uniformly from rest down a steady hill inclined at 15 degrees to the horizontal[...] - Work out the energy lost to friction after the car has traveled 100m - The average...