bead rolling down a frictionless track, help please?

Question

Speed and net work of bead down a frictionless track?

A bead of mass m = 5.10 kg is released from point A and slides on the frictionless track shown in Figure P5.30. The height of A is ha = 5.90 m.

Figure P5.30: http://i165.photobucket.com/albums/u78/RLB31384/p5-30alt.gif
(a) Determine the bead's speed at points B and C. point B m/s
point C m/s
(b) Determine the net work done by the force of gravity in moving the bead from A to C.

Answer 1

Conservation of energy and conversion of potential energy to kinetic energy are the keys to this problem.

Potential energy at point A = m*g*ha (ha = 5.9m)
Kinetic energy at point A = 0

Potential energy at point B = m*g*hb (hb = 3.2m)
Kinetic energy at point B = 0.5*m*vb^2

The change in potential energy from point A to point B will be equal to the kinetic energy at point B:

m*g*ha - m*g*hb = 0.5*m*vb^2 solve for vb

Potential energy at point C = m*g*hc (hc = 2m)
Kinetic energy at point C = 0.5*m*vc^2

The change in potential energy from point A to point C will be equal to the kinetic energy at point C:

m*g*ha - m*g*hc = 0.5*m*vc^2 solve for vc

The work done from point A to point C is the difference in the potential energies at that point.