Speed and net work of bead down a frictionless track?

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

using conservation of energy

.5*m*v^2=m*g*h

v=sqrt(2*g*h)

For point B
v=sqrt(2*9.8*(5.9-3.2))

=7.27

For point C
v=sqrt(2*9.8*(5.9-2))
=8.74

Work = F*d

The force is m*a
w=5.1*9.8*(5.9-2)
=195 J

j