Respuesta :
Spring constant k = 75 N/m
Hooke's law for this spring F(x) = 75x
Work = 1.5 J
What is work?
- When an object is moved over a distance by an external force, at least a portion of that force must be applied in the direction of the displacement. This is known as work.
- The amount of work required to move an object down a line from point a to point b in the direction of a variable force F is
[tex]$W=\int_{a}^{b} F(x) d x$[/tex]
First solution
According to Hooke's law, the spring must be held stretched at x=0.2m by a force of 15N.
F (0.2) = k (0.2 m) = 15 N
Solve for k
k = 75 N/m
Hence, Hooke's law is F(x) = 75x
Second solution
The spring is compressed from x = 0 to X = 0.2,
[tex]$W=\int_{0}^{0.2} 75 x d x=\left.\left(\frac{75 x^{2}}{2}\right)\right|_{0} ^{0.2}$[/tex]
W = 1.5 J
So, the work is required to compress the spring 0.2 m from its equilibrium position W = 1.5 J
Learn more about work here:
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