Question

a spring has a constant of k=7.50 n/m. if the spring is displaced 0.550m from its equilibrium position, what is the force that the spring exerts? assume for this and for all other questions in the pre-laboratory that g=9.8m/s^2. show your work

Answer 1

Answer:

$F=4.125\ N$

Explanation:

Given:

• spring constant, $k=7.5\ N.m^{-1}$

• displacement in the equilibrium position, $\delta x=0.55\ m$

We know the force exerted by a strained spring is given as:

$F=k.\delta x$

$F=7.5\times 0.55$

$F=4.125\ N$

Answer 2

Answer:F=4.125 N

Explanation:

Given

Spring constant $k=7.50\ N/m$

Displacement $x=0.55\ m$

Spring Force is given  by the Product of spring constant and displacement of spring from mean position

$F=k\cdot x$

$F=7.50\cdot 0.55$

$F=4.125\ N$