A spring with a mass block at the end of it and with a stiffness of 8 $N/m$ and a relaxed length of 20 $cm$ is attached to a chamber wall that results in its oscillations being horizontal. At a particular time the location of the block mass is $\langle .38,0,0 \rangle\,m$ relative to an origin point where the spring is attached to the chamber wall. Determine the force exerted by the spring on the mass at this instant.

• Spring has relaxed length of 0.2m, $L_0=0.2\,m$
• Spring has spring constant of $8 N/m$, $k_s=8\,N/m$
• At the moment of interest, the mass block is at position $\vec{L} = \langle .38,0,0 \rangle m$
• The net force acting on system is due to spring force (the gravitational force exerted by the Earth has the same magnitude as the force exerted by the horizontal surface)

• The force that the spring exerts

• Origin is at chamber wall $\langle 0,0,0 \rangle\,m$
• Assume no forces due to drag or to friction

$ {\vec F_{spring}} = -k_s\vec{s}$

$ |\vec{s}| = |L - L_0|$

To determine the spring force, you will need to compute: $$ {\vec F_{spring}} = -k_s\vec{s} = -k_s|\vec{s}|\hat{s}$$

You will start be determining the position vector ($\vec{L}$) of the mass and the length of the position vector ($|\vec{L}|$), $$\vec{L} = \langle 0.38,0,0 \rangle m - \langle 0,0,0 \rangle m = \langle 0.38,0,0 \rangle m$$

$$|\vec{L}| = 0.38m$$

These can be used to compute the unit (direction) vector for the stretch ($\hat{s}$), which is in the same direction as the position vector: $$\hat{s} = \hat{L} = \dfrac{\langle 0.38,0,0\rangle}{0.38} = \langle 1,0,0 \rangle$$

You can then compute the magnitude of the stretch $(|\vec{s}|)$: $$ |\vec{s}| = |L - L_0| = 0.38m - 0.20m = 0.18m$$

Finally, you can compute the force:

$$\vec{F} = -k_s|\vec{s}|\hat{s} = -(8N/m)(0.18m)\langle 1,0,0\rangle = \langle -1.44,0,0 \rangle\,N$$

which points to the left. That is consistent with the diagram above.