A spring of length 0.4 m, fixed at one end in the vertical direction is extended by 5 cm, by attaching a mass of 2 kg. Then it is further pulled by 5 cm and left to itself. What is the final potential energy ? What is the amplitude of oscillation ? `(g =10 m//s^(2))`

Options

A. 2 J, 10 cm
B. 1 J, 10 cm
C. 2 J, 5 cm
D. 3 J, 10 cm

Answer

C

Explanation

`F=2xx10=20 N because F=Kx`
`therefore K=(F)/(x)=(20)/(0.05)=400 N//m`
Total extension =5+5=10 cm =0.1 m
`therefore` Final `P.E. =(1)/(2)Kx^(2)=(1)/(2)xx400xx((1)/(10))^(2)=2J`
The amplitude of oscillation is only 5 cm (and not 10 cm). 5 cm is the equilibrium position.