Exercises — Hooke's law — spring force F = −kx

WHAT we do: put the numbers into $F=-kx$. WHY: the question gives us $k$ and $x$ directly — this is the raw definition. $F=-(150)(0.04)=-6\text{N}$ The minus tells us the direction (spring pulls back toward $x=0$, i.e. to the left of a rightward stretch). The question asked only for the magnitude, so we drop the sign: $|F|=k|x|=\boxed{6\text{N}}$.

WHAT: substitute the negative $x$. Because the question asks for the sign and direction, we keep the full vector form $F=-kx$. $F=-kx=-(500)(-0.03)=+15\text{N}$ WHY it's positive: we compressed the spring to the left ($x<0$), so it shoves back to the right ($F>0$). The two minus signs (one in the formula, one in $x$) cancel. This is the whole point of the minus sign: it makes $F$ flip whenever $x$ flips. Answer: $\boxed{+15\text{N, pointing right (back toward equilibrium)}}$.

WHAT: use $U=\frac{1}{2}k{x}^2$. WHY squared: energy depends on $x^2$, so the sign of $x$ doesn't matter — a stretch and a squeeze of the same size store the same energy. $U=\frac{1}{2}(500)(0.03{)}^2=\frac{1}{2}(500)(0.0009)=0.225\text{J}$ Answer: $\boxed{0.225\text{J}}$. Energy is never negative here.

WHAT: at rest, the spring's upward pull balances gravity's downward pull. WHY: "at rest" means net force is zero, so the restoring force magnitude $kx$ equals the weight $mg$. We work with magnitudes here because both forces are equal in size and opposite in direction — the signs already cancel in the balance. $kx=mg\Rightarrow k=\frac{mg}{x}=\frac{(0.8)(9.8)}{0.16}=49\text{N/m}$ Answer: $\boxed{49\text{N/m}}$. See Conservation of mechanical energy later for what happens if we drop it.

WHAT: invert $U=\frac{1}{2}k{x}^2$ for $x$. WHY: we know the energy and stiffness, want the displacement — so we solve for $x$. $x^2=\frac{2U}{k}=\frac{2(2)}{250}=0.016\Rightarrow x=\sqrt{0.016}=0.1265\text{m}$ Answer: $\boxed{x\approx 0.126\text{m}}$ (about 12.6 cm). We took the positive root because "how far you stretch" is a length.

WHAT: find $k$ from the first data point, then use it for the second. WHY / sign note: both stretches are positive lengths and the question asks only for the size of the force, so we use the magnitude form $|F|=kx$ throughout (the minus sign in $F=-kx$ only tells us the pull is toward equilibrium — the direction, which we're not asked for here). $k=\frac{|F|}{x}=\frac{18}{0.09}=200\text{N/m}$ $|F|=kx=(200)(0.15)=30\text{N}$ Answer: $\boxed{30\text{N}}$. (Notice $|F|\propto x$: $x$ went up by $\frac{0.15}{0.09}=1.67\times$, and so did $|F|$: $\frac{30}{18}=1.67$.)

WHAT: the energy added is the difference of the two $U$ values (the shaded strip in the figure). WHY subtract: stored energy is the area under the line from $0$ to $x$. Going from $x_1$ to $x_2$ adds only the strip between them. $U_2=\frac{1}{2}(400)(0.10{)}^2=2.0\text{J},U_1=\frac{1}{2}(400)(0.05{)}^2=0.5\text{J}$ $\Delta U=U_2-U_1=2.0-0.5=1.5\text{J}$ The first $0.05\text{m}$ stored only $0.5\text{J}$, but the second (equal-length) stretch stored $1.5\text{J}$ — three times more. Answer: $\boxed{1.5\text{J}}$. The lesson: later stretches are far more expensive, because the force you fight is already large. See Work done by a variable force.

WHAT: solve $x=\sqrt{2U/k}$ for each spring. WHY: equal energy but different stiffness — the softer spring must move further to bank the same energy. $x_A=\sqrt{\frac{2(1.5)}{300}}=\sqrt{0.01}=0.10\text{m}$ $x_B=\sqrt{\frac{2(1.5)}{100}}=\sqrt{0.03}=0.1732\text{m}$ The softer spring B stretches further: $\boxed{x_A=0.10\text{m},x_B\approx 0.173\text{m}}$. Since $x\propto 1/\sqrt{k}$, cutting $k$ by a factor 3 stretches it $\sqrt{3}\approx 1.73\times$ more.

WHAT: all the spring's stored energy becomes the block's kinetic energy. WHY: frictionless ⇒ mechanical energy is conserved. The spring's $U$ empties into $\frac{1}{2}m{v}^2$. See Conservation of mechanical energy and Elastic potential energy. $\underset{\text{spring PE}}{\underbrace{\frac{1}{2}k{x}^2}}=\underset{\text{block KE}}{\underbrace{\frac{1}{2}m{v}^2}}$ $\frac{1}{2}(800)(0.05{)}^2=\frac{1}{2}(0.2)v^2$ $1.0=0.1v^2\Rightarrow v^2=10\Rightarrow v=\sqrt{10}=3.162\text{m/s}$ Answer: $\boxed{v\approx 3.16\text{m/s}}$.

WHAT: at half speed the block has $\frac{1}{4}$ of its max kinetic energy (since KE $\propto v^2$), so $\frac{3}{4}$ of the energy is still in the spring. WHY: total energy $E=\frac{1}{2}k{x}_0^2=1.0\text{J}$ is fixed. If KE $=\frac{1}{4}E$, then spring PE $=\frac{3}{4}E$. $\frac{1}{2}k{x}^2=\frac{3}{4}E=\frac{3}{4}(1.0)=0.75\text{J}$ $x^2=\frac{2(0.75)}{800}=0.001875\Rightarrow x=0.0433\text{m}$ Answer: $\boxed{x\approx 0.0433\text{m}}$. Note it's not halfway ($0.025$ m) — because energy depends on $x^2$, the spring gives up its energy fastest near the end.

Because $F=-kx$ is exactly the restoring-force law, the mass performs Simple Harmonic Motion.

(a) WHAT/WHY: $\omega =\sqrt{k/m}$ comes from setting $ma=-kx$ (Newton's law with the spring force). It's the "rate of swinging," in rad/s. $\omega =\sqrt{\frac{200}{0.5}}=\sqrt{400}=20\text{rad/s}$

(b) WHAT/WHY: max speed happens at $x=0$, where all energy is kinetic. Set $\frac{1}{2}k{A}^2=\frac{1}{2}m{v}_{\max }^2$ with amplitude $A=0.10\text{m}$. $v_{\max }=A\sqrt{\frac{k}{m}}=A\omega =(0.10)(20)=2\text{m/s}$

(c) WHAT/WHY: use energy conservation — total energy splits between spring PE at $x=0.06$ and KE. $\frac{1}{2}k{A}^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$ $\frac{1}{2}(200)(0.10{)}^2=\frac{1}{2}(200)(0.06{)}^2+\frac{1}{2}(0.5)v^2$ $1.0=0.36+0.25v^2\Rightarrow v^2=\frac{0.64}{0.25}=2.56\Rightarrow v=1.6\text{m/s}$ Answers: $\boxed{\omega =20\text{rad/s},v_{\max }=2\text{m/s},v(0.06)=1.6\text{m/s}}$.

Here $x\equiv r-r_0$ is exactly the "displacement from natural length" of the parent note, only for a bond instead of a coil. This is where the tool $F=-\frac{dU}{dr}$ from Conservative forces and potential energy finally does real work.

(a) WHAT/WHY — apply $F=-dU/dr$ and evaluate at $r_0$. Differentiate the energy. Writing $s=r_0/r$ for brevity, use the chain rule $\frac{d}{dr}(r_0/r{)}^n=-\frac{n}{r}(r_0/r{)}^n$: $\frac{dU}{dr}=U_0\left[-\frac{12}{r}(r_0/r{)}^{12}+\frac{12}{r}(r_0/r{)}^6\right]=\frac{12U_0}{r}\left[(r_0/r{)}^6-(r_0/r{)}^{12}\right]$ $F=-\frac{dU}{dr}=\frac{12U_0}{r}\left[(r_0/r{)}^{12}-(r_0/r{)}^6\right]$ At $r=r_0$ we have $r_0/r=1$, so the bracket is $1-1=0$. The force vanishes at $r_0$ — confirming it is the equilibrium (bottom of the well), exactly like $x=0$ for a spring. ✓

(b) WHAT/WHY — the stiffness is the curvature $U^{\prime \prime }(r_0)$. Just as in the parent note's Taylor argument, the bottom of any energy well looks like $U\approx \frac{1}{2}k{x}^2$, whose curvature is $\frac{d^{2}U}{d{x}^2}=k$. Since $x=r-r_0$ shifts by a constant, $\frac{d^{2}U}{d{x}^2}=\frac{d^{2}U}{d{r}^2}$, so $k=U^{\prime \prime }(r_0)$. Differentiate $dU/dr$ once more (again using $\frac{d}{dr}(r_0/r{)}^n=-\frac{n}{r}(r_0/r{)}^n$ term by term): $\frac{d^{2}U}{d{r}^2}=U_0\left[\frac{12\cdot 13}{r^2}(r_0/r{)}^{12}-\frac{6\cdot 7}{r^2}(r_0/r{)}^6\right]=\frac{U_0}{r^2}\left[156(r_0/r{)}^{12}-42(r_0/r{)}^6\right]$ At $r=r_0$ the ratios are $1$, so: $k=U^{\prime \prime }(r_0)=\frac{U_0}{r_0^2}[156-42]=\frac{114U_0}{r_0^2}?$ Wait — the two exponents are $12$ and $6$, but the coefficient in front of the second-derivative must come from differentiating the force, not the raw energy powers. Redo cleanly from $F$: differentiate $F(r)=\frac{12U_0}{r}\left[(r_0/r{)}^{12}-(r_0/r{)}^6\right]$ and evaluate the slope of the force at $r_0$, since $k=-\frac{dF}{dr}{|}_{r_0}$. At $r=r_0$ the leading $1/r$ factor is constant to first order and the surviving derivative acts on the bracket, giving $\frac{dF}{dr}{|}_{r_0}=\frac{12U_0}{r_0}\cdot \frac{1}{r_0}[-12+6]=-\frac{72U_0}{r_0^2}$. Therefore $k=-\frac{dF}{dr}{|}_{r_0}=\frac{72U_0}{r_0^2}.$ The factor 72 is just $12\times 6$ — the product of the two exponents in the Lennard-Jones energy. ✓

(c) WHAT/WHY — plug in numbers. $k=\frac{72(1.6\times 10^{-19})}{(3.0\times 10^{-10}{)}^2}=\frac{1.152\times 10^{-17}}{9.0\times 10^{-20}}=128\text{N/m}$ Answer: $\boxed{k\approx 128\text{N/m}}$. Astonishingly, a single molecular bond has a stiffness comparable to a small lab spring — because at that scale the forces are enormous relative to the tiny displacements involved. This is precisely why Interatomic forces near equilibrium reproduce Hooke's law, and why solids ring, vibrate, and store elastic energy just like macroscopic springs.