Question

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

Answer 1

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

Formula:
$F = k*x$

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

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

Formula:
$E = \frac{k*x^2}{2}$

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