A physics problem. A person riding a bike has a combined weight of 100kg.?

Question

He is going 8 m/s on a straight plain when he comes to a hill. The hill is 6 meters tall. What is the velocity of the person and the bike when they reach the bottom of the hill. Don't take air resistance into consideration.

There isnt enough information to do the problem.
I dont understand how you are supposed to get the length of the hill (x) or the time it took to get down the hill.

Can someone help me?

Answer 1

This is an energy problem. The person/bike system has potential and kinetic energy. The total energy at the top of the hill = the total energy at the bottom. At the top the system has both KE and PE but at the bottom the system only has KE
You do not need to get the length of the hill or the time.

The KE = (1/2) mv^2
The PE = mgh

Energy at top = Energy at bottom

PE + KE = KE
(100)(9.8)(6) + (1/2)(100)(64) = (1/2)(100) v^2

9 080 / 50 = v^2

181.6 = v^2

v = 13.5 m/s

Answer 2

No, there *is* enough information!

It doesn't matter how far the person travels horizontally, because you don't need to do any work to move at 90 degrees to the direction in which gravity is pulling. (Well, there's air resistance to think about, but the question specifically says to ignore that). And it doesn't matter how long it takes for them to get there.

The only that matters is that the cyclist has converted some potential energy to kinetic energy, by travelling downhill.

You first need to work out the cyclist's kinetic energy at 8 m/s before descending the hill (.5 * m * v ** 2). Then you need to work out how much potential energy they lost descending 6 m (m * g * h). Since none of that energy has gone anywhere else, it must all have been converted to kinetic energy. So, add up the initial KE to the amount of KE converted from PE to find the cyclist's KE at the bottom of the hill. Then you need to work out the speed they were travelling.

KE at top of hill = .5 * 100 * 8 * 8 = 3200 J
PE lost in descent = 100 * 10 * 6 = 6000 J (assuming g = 10 m/s2)
KE at bottom of hill = 9200 J
v ** 2 = 9200 / (.5 * 100) = 184
so v = 13.56 m/s (to 4 significant figures).

Answer 3

I think the purpose of this problem is to get you to think in terms of energy conservation, but there is a lot that is left out in the description of the statement. For instance, are we to assume the person stops pedalling on the slope? If that is not the case, then there is little we can do without more information. if that is the case, are they assuming there is no friction between the bike and the hill? If they are assuming that, I can tell they haven't biked up a hill in a while.

So, if we make all the simplifying assumptions they seem to want us to make, i.e., no friction and coasting up and down the hill, then the conservation of energy tells you that the amount of energy you have doesnt change. At the base of the hill, the biker had some kinetic energy (1/2 mv^2) and no potential energy. At the base on the opposite side, he must have the same kinetic energy (since there is no potential difference at the same elevation), and if the kinetic energy is the same, the speed is the same. But this is not a realistic problem at all. Good luck and hope this helps.