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a 60 - kilogram person stands on a compression spring. the spring const…

Question

a 60 - kilogram person stands on a compression spring. the spring constant is 2500 newtons per meter. what is the displacement of the spring? 0.024 m 0.1176 m 0.2352 m 0.00392 m

Explanation:

Step1: Calculate the force exerted by the person

The force exerted by the person is equal to their weight. Using the formula $F = mg$, where $m = 60\ kg$ and $g= 9.8\ m/s^{2}$. So $F=60\times9.8 = 588\ N$.

Step2: Use Hooke's law to find displacement

Hooke's law is $F = kx$, where $F$ is the force, $k$ is the spring - constant, and $x$ is the displacement. We know $F = 588\ N$ and $k = 2500\ N/m$. Rearranging for $x$, we get $x=\frac{F}{k}$. Substituting the values, $x=\frac{588}{2500}=0.2352\ m$.

Answer:

$0.2352\ m$