Question

2) A 2.0 g bead slides along a frictionless wire, as shown in the figure. At point 4. the bead is moving

Answer 1

Answer:

(a) $2\times 10^{-2} J$

(b)  $2\times 10^{-2} J$

(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,$9.81 m/s^{2}$ for g and 1 m for h we obtain

$PE=0.002 kg\times 9.81 m/s^{2}\times 1m=0.01962 J\approx 2\times 10^{-2} J$

(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 $v=\sqrt {2gh}$

$v=\sqrt {2*9.81*1}=4.429446918\approx 4.43 m/s$

$KE=0.5*0.002*4.43^{2}=0.01962 J\approx 2\times 10^{-2} J$

(c)

As already illustrated in part b

$v=\sqrt {2gh}$

$v=\sqrt {2*9.81*1}=4.429446918\approx 4.43 m/s$

(d)

From the law of conservation of energy

Energy at point A equals energy at point C

$v=\sqrt {2g\triangle h}$

$v=\sqrt {2*9.81*(1-0.8)}=1.980908882 \approx 2 m/s$