2) a 2.0 g bead slides along a frictionless wire, as shown in the figure. at point 4. the bead is moving
to the right but with negligible speed.
jeg
le
10) cm
hi
80 cm
+-- - - - - potential energy = 0
(a) what is the potential energy of the bead at point 4?
(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 om

(a)

(b)  

(c) 4.43 m/s

(d) 2 m/s

Explanation:

Using the attached image, point 4 is point A

Since potential energy PE=mgh where m is mass, g is acceleration due to gravity and h is height.

The height is 100cm equivalent to 1 m

Substituting 2g equivalent to 0.002 Kg for m, for g and 1 m for h we obtain

(b)

Kinetic energy is given by

KE=0.5mv^{2} where v is the velocity, m is mass and KE is kinetic energy

Substituting m for 0.002 Kg and

(c)

As already illustrated in part b

(d)

From the law of conservation of energy

Energy at point A equals energy at point C