Answer:
(a) [tex]2\times 10^{-2} J[/tex]
(b) [tex]2\times 10^{-2} J[/tex]
(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,[tex]9.81 m/s^{2}[/tex] for g and 1 m for h we obtain
[tex]PE=0.002 kg\times 9.81 m/s^{2}\times 1m=0.01962 J\approx 2\times 10^{-2} J[/tex]
(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 [tex]v=\sqrt {2gh}[/tex]
[tex]v=\sqrt {2*9.81*1}=4.429446918\approx 4.43 m/s[/tex]
[tex]KE=0.5*0.002*4.43^{2}=0.01962 J\approx 2\times 10^{-2} J[/tex]
(c)
As already illustrated in part b
[tex]v=\sqrt {2gh}[/tex]
[tex]v=\sqrt {2*9.81*1}=4.429446918\approx 4.43 m/s[/tex]
(d)
From the law of conservation of energy
Energy at point A equals energy at point C
[tex]v=\sqrt {2g\triangle h}[/tex]
[tex]v=\sqrt {2*9.81*(1-0.8)}=1.980908882
\approx 2 m/s[/tex]
