Answer:
a.6.5025 J
b.6.5025 J
Explanation:
We are given that
Mass of pellet,m=0.27 g=[tex]0.27\times 10^{-3} kg[/tex]
1 kg=1000 g
Spring constant,k=1800 N/m
x=8.5 cm=[tex]8.5\times 10^{-2} m[/tex]
1m=100 cm
a.Potential energy stored in the compressed spring is given by
P.E=[tex]\frac{1}{2}kx^2[/tex]
[tex]P.E=\frac{1}{2}(1800)(8.5\times 10^{-2})^2[/tex]
[tex]P.E=6.5025 J[/tex]
b.By using law of conservation of energy
P.E of spring=K.E of the pellet
K.E of the pellet=6.5025 J