Some Particular Forces — Question 10

Question 10

A spring force has a signed displacement law

A horizontal spring exerts $F_x=-kx$ on a 2 kg slider, with $k=50$ N/m and x measured from the relaxed position. The track is frictionless and vertical forces balance.

1. Find force and acceleration at $x=+0.10$ m and $x=-0.06$ m.

2. At $x=0$, must the slider be at rest?

3. Explain why the spring is not interchangeable with a tension-only cord.

Question 10 - Solution

A spring force has a signed displacement law

(a) Apply the restoring sign.

$$F_x=-kx,\quad a_x=-\frac{k}{m}x:\quad\boxed{(F,a)_{.10}=(-5,-2.5),\quad(F,a)_{-.06}=(3,1.5)}\ \text{(SI)}$$ At +0.10 m, $F_x=-5$ N and $a_x=-2.5$ m/s$^2$. At -0.06 m, $F_x=+3$ N and $a_x=+1.5$ m/s$^2$.

(b) Distinguish force from velocity.

$$x=0\ \Rightarrow\ F_x=0,\ a_x=0,\quad v_x\text{ unrestricted by this instantaneous condition}$$ The spring force and horizontal acceleration vanish at x=0, but the slider may pass through with nonzero velocity. The instantaneous force law does not specify the motion’s initial conditions.

(c) Compare physical models.

$$F_xx=-kx^2\le0,\quad F_x\text{ changes sign at }x=0;\qquad T_{\mathrm{cord}}\ge0$$ The supplied spring law allows both pulling and pushing relative to the relaxed position. A flexible cord can sustain tension but not a compressive push. No full oscillation solution is needed to distinguish these models.