Can Anyone Help Me with Physics?

Question

I am confused on this problem. Any input would be helpful :)

A bead of mass m = 7.60 kg is released from point A and slides on the frictionless track shown in the figure linked below. The height of A is ha = 7.90 m.

Figure: http://www.webassign.net/sercp/p5-30alt....

(a) Determine the bead's speed at points B and C (in m/s).
(b) Determine the net work done by the force of gravity in moving the bead from A to C (in J).

http://www.webassign.net/sercp/p5-30alt.gif

It should be .gif after the alt.

Thanks for all of your time Wicked Mickey :)

Answer 1

This is a bit of a trick question. One of the things you learn here is there is a difference between velocity and speed. Velocity is speed in a direction...

The only force acting on the bead is gravity so

v2=v02+2a(x-x0)
v2=0-2*(9.81m/s2)(3.29m-7.9m)
v2=-19.62m/s2(-4.61m)
v2=90.45 m2/s2
v=9.51 m/s

likewise speed at point c is 10.76 m/s

The work is then Force times the distance. Force = mass * acceleration.

The mass is 7.60 kg. The acceleration is -9.81 m/s2 The distance from a to b is -4.61 m, from a to c is -5.9 m

So work from a to b is 343.7 kg m/s2 m or 343.7 Nm or 343.7 Joules.

work from a to c is 439.9 kg m/s2 m or 439.9 Nm or 439.9 Joules.

Answer 2

I won't do all the math but I'll tell you how to do it.
You know that the total energy equals to the sum of the kinetic and the potencial energys(Et=Ek+Ep). And that the kinetic energy is the product between the mass, the gravity and the height(Ek=m/2v^2). And that the kinetic energy is half of the product of the mass with the square of the velocity(Ep=mgh). You also must remember that the total energy is always the same. In the point A you have to find the total energy, and you can see that it only has Ep. Then put the information into the formula. In B, you know it has both kinds of energys, then use the information (remember that you already have one side of the equation (Et=Ek+Ep) that is the total energy you found in A.
Well.....I'm not really sure if this is how it's done...because I studied it 4 years ago and never saw it again! But I guess that is it...put the info you know and give it a try.
as english is not my language, i dont know what is the second question asking....but you can go for there.