Respuesta :
Answer:
The value of spring constant is 266.01 [tex]\frac{N}{m}[/tex]
Explanation:
Given:
Mass of pellet [tex]m = 2.01 \times 10^{-2}[/tex] kg
Height difference of pellet rise [tex]h_{f} - h_{o} = 6.03[/tex] m
Spring compression [tex]x = 9.45 \times 10^{-2}[/tex] m
From energy conservation law,
Spring potential energy is stored into potential energy,
[tex]mg(h_{f} -h_{o}) = \frac{1}{2} kx^{2}[/tex]
Where [tex]k =[/tex] spring constant, [tex]g = 9.8 \frac{m}{s^{2} }[/tex]
[tex]k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }[/tex]
[tex]k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }[/tex]
[tex]k = 266.01[/tex] [tex]\frac{N}{m}[/tex]
Therefore, the value of spring constant is 266.01 [tex]\frac{N}{m}[/tex]
