Question

A 1.8 g bead slides along a wire, as shown in the figure. At point A, the bead is at rest. Neglect friction.
(a) What is the potential energy of the bead at point A?
(b) What is the kinetic energy of the bead at point B?
(c) What is the speed of the bead at point B?
(d) What is the speed of the bead at point C?

Answer 1

Explanation:

a) Potential energy at point A :

P.E = mgh

P.E = (0.0018) x 9.8 x 0.1

P.E = 0.001764‬ Joules

b) AT point B, all the potential energy will be converted to kinetic energy so kinetic energy at point B ;

K.E =  0.001764‬ Joules

c) Kinetic energy = 1/2 m x v^2

    0.001764 = 1/2 x 0.0018 x v^2

   v^2 = 1.96

   v = 1.4 m/s

d) Potential energy at point C would be;

   P.E = (0.0018) x 9.8 x 0.08

   P.E = 0.0014112‬ Joules

   

  Kinetic energy at point C = Total energy - P.E

 K.E = 0.001764 - 0.0014112

 K.E = 0.0003528

 0.0003528 = 1/2 x 0.0018 x v^2

 v = 0.626 m/s