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A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?

Respuesta :

dexteright02
Data:
F (force) = ? (Newton)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
[tex]F = k*x[/tex]

Solving: 
[tex]F = k*x[/tex]
[tex]F = 50*0.15[/tex]
[tex]\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark[/tex]

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
[tex]E = \frac{k*x^2}{2} [/tex]

Solving:(Energy associated with this stretching)
[tex]E = \frac{k*x^2}{2} [/tex]
[tex]E = \frac{50*0.15^2}{2} [/tex]
[tex]E = \frac{50*0.0225}{2} [/tex]
[tex]E = \frac{1.125}{2} [/tex]
[tex]\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark[/tex]

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