The 10 N mark must be at a distance of 6.67 cm from the 0 N mark
Explanation:
We can solve this problem by using Hooke's law.
In fact, the restoring force in a spring is given by (in magnitude)
[tex]F=kx[/tex]
where
F is the force applied
k is the spring constant
x is the elongation of the spring
For the spring in this problem, we have
k = 150 N/m
F = 10.0 N (force applied)
Solving the equation for x, we find
[tex]x=\frac{F}{k}=\frac{10}{150}=0.0667 m[/tex]
So, the 10 N mark must be at 6.67 cm from the 0.0 N mark.
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