Exercises — Spring potential energy — derivation

What: plug straight into $U=\frac{1}{2}k{x}^2$. Why: the spring is ideal, $x$ is already measured from the relaxed length, so no adjustment is needed. $U=\frac{1}{2}(300)(0.05{)}^2=\frac{1}{2}(300)(0.0025)=0.375\text{J}$ Answer: $U=0.375\text{J}$.

What: compare $U$ for $x=+0.05$ and $x=-0.05$. Why: $x$ is squared in $U=\frac{1}{2}k{x}^2$, so the sign disappears. Both give $(0.05{)}^2=0.0025$. $U=\frac{1}{2}(300)(-0.05{)}^2=0.375\text{J}$ Answer: exactly the same, $0.375\text{J}$. Compression and stretch of equal size store equal energy.

What: rearrange $U=\frac{1}{2}k{x}^2$ to solve for $k$. Why: we know $U$ and $x$, want $k$ — isolate it. $k=\frac{2U}{x^2}=\frac{2(2)}{(0.20{)}^2}=\frac{4}{0.04}=100\text{N/m}$ Answer: $k=100\text{N/m}$.

What: compute both energies and take their ratio. Why: because $U\propto x^2$, tripling $x$ should scale $U$ by $3^2$. Let's verify. $U_1=\frac{1}{2}(250)(0.10{)}^2=1.25\text{J}$ $U_2=\frac{1}{2}(250)(0.30{)}^2=11.25\text{J}$ $\frac{U_2}{U_1}=\frac{11.25}{1.25}=9$ Answer: energy grows 9× (not 3×), because $U\propto x^2$.

What: the extra work is the difference of stored energies. Why: work done by you against a conservative spring = change in stored PE. We already have both from Q1. $W=U_2-U_1=11.25-1.25=10\text{J}$ Answer: $W=10\text{J}$. What it looks like: on the force–$x$ graph this is the trapezoid strip between $x=0.10$ and $x=0.30$ — the big triangle minus the small triangle.

What: set stored spring energy equal to the block's kinetic energy. Why: Conservation of mechanical energy — no friction, so all elastic PE becomes motion energy $\frac{1}{2}m{v}^2$. $U=\frac{1}{2}(800)(0.05{)}^2=\frac{1}{2}(800)(0.0025)=1\text{J}$ $\frac{1}{2}m{v}^2=1\Rightarrow \frac{1}{2}(0.4)v^2=1\Rightarrow v^2=\frac{1}{0.2}=5$ $v=\sqrt{5}\approx 2.236\text{m/s}$ Answer: $v\approx 2.24\text{m/s}$.

What (part 1): at rest the spring force balances gravity: $kx=mg$. Why: "rests" means zero net force, so the upward spring pull equals the downward weight. $x=\frac{mg}{k}=\frac{(0.5)(10)}{200}=\frac{5}{200}=0.025\text{m}$ What (part 2): stored spring energy at that stretch: $U=\frac{1}{2}k{x}^2=\frac{1}{2}(200)(0.025{)}^2=\frac{1}{2}(200)(0.000625)=0.0625\text{J}$ Answer: stretch $=0.025\text{m}$, stored energy $U=0.0625\text{J}$. Note: we contrast with Gravitational potential energy $U=mgh$ (constant force) — here the spring force changes, so it needs the triangle formula.

What: compute triangle area = $\frac{1}{2}\times \text{base}\times \text{height}$ with base $x$ and height $F=kx$. Why: the area under a force–displacement graph is the work done (energy stored).

• base $=x=0.025\text{m}$
• height $=kx=200(0.025)=5\text{N}$ $\text{area}=\frac{1}{2}(0.025)(5)=0.0625\text{J}$ Answer: matches Q1 exactly — $0.0625\text{J}$. See the figure below: the height is the tip of the triangle, the base is the stretch.

What: released energy $=U_{\text{start}}-U_{\text{end}}$. Why: the spring gives up the difference in stored energy as it partially relaxes. $U_{\text{start}}=\frac{1}{2}(1000)(0.10{)}^2=5\text{J}$ $U_{\text{end}}=\frac{1}{2}(1000)(0.04{)}^2=0.8\text{J}$ $\Delta U=5-0.8=4.2\text{J}$ Answer: $4.2\text{J}$ released.

What: all spring PE converts to gravitational PE at the highest point (where speed $=0$). Why: Conservation of mechanical energy — frictionless, so spring energy → gravitational energy $mgh$. $\frac{1}{2}k{x}^2=mgh$ $\frac{1}{2}(400)(0.20{)}^2=(0.5)(10)h$ $\frac{1}{2}(400)(0.04)=5h\Rightarrow 8=5h\Rightarrow h=1.6\text{m}$ Answer: $h=1.6\text{m}$ (measured as vertical rise, independent of ramp angle).

What: spring energy = kinetic energy gained + energy lost to friction. Why: energy bookkeeping — friction removes $f\times d$ of energy as heat; the rest is motion. $U=\frac{1}{2}(400)(0.20{)}^2=8\text{J}$ $E_{\text{friction}}=fd=2\times 1.5=3\text{J}$ $\frac{1}{2}m{v}^2=U-E_{\text{friction}}=8-3=5\text{J}$ $\frac{1}{2}(0.5)v^2=5\Rightarrow v^2=\frac{5}{0.25}=20\Rightarrow v=\sqrt{20}\approx 4.472\text{m/s}$ Answer: $v\approx 4.47\text{m/s}$.

What: find where net force on the block is zero. Why: speed keeps increasing while the spring still pushes ($F>0$) and stops increasing once the push ends. Maximum speed is exactly where the spring force reaches zero — the natural length, $x=0$. At $x=0$ all the stored energy $U=\frac{1}{2}(800)(0.05{)}^2=1\text{J}$ has become kinetic: $\frac{1}{2}(0.4)v_{\max }^2=1\Rightarrow v_{\max }^2=5\Rightarrow v_{\max }=\sqrt{5}\approx 2.236\text{m/s}$ Answer: maximum speed occurs at the relaxed length ($x=0$), $v_{\max }\approx 2.24\text{m/s}$.

What: the applied force balances the spring: $F_{\text{applied}}=k{x}^{\prime }+\beta x^{\prime 3}$. Integrate it from $0$ to $x$. Why: the derivation logic from the parent note is general — Work done by a variable force means $U={\int }_0^x{F}_{\text{applied}}d{x}^{\prime }$ whatever the force law, as long as it's conservative. $U={\int }_0^x(k{x}^{\prime }+\beta x^{\prime 3})d{x}^{\prime }=k\frac{x^2}{2}+\beta \frac{x^4}{4}=\frac{1}{2}k{x}^2+\frac{1}{4}\beta x^4$ Now evaluate: $U=\frac{1}{2}(100)(0.10{)}^2+\frac{1}{4}(5000)(0.10{)}^4$ $=\frac{1}{2}(100)(0.01)+\frac{1}{4}(5000)(0.0001)$ $=0.5+0.125=0.625\text{J}$ Answer: $U(x)=\frac{1}{2}k{x}^2+\frac{1}{4}\beta x^4$; numerically $0.625\text{J}$. Insight: the linear part is our familiar triangle; the cubic adds a curved sliver on top because the force curls up faster than a straight line.

What (step 1): find the effective spring constant. In series, $\frac{1}{k_{\text{eff}}}=\frac{1}{k_1}+\frac{1}{k_2}$. Why: the same tension $F$ stretches each ($F=k_1{x}_1=k_2{x}_2$), and total extension $x=x_1+x_2$. Substituting gives the reciprocal rule. $\frac{1}{k_{\text{eff}}}=\frac{1}{300}+\frac{1}{600}=\frac{2}{600}+\frac{1}{600}=\frac{3}{600}=\frac{1}{200}$ $k_{\text{eff}}=200\text{N/m}$ What (step 2): total energy = $\frac{1}{2}{k}_{\text{eff}}{x}^2$. Why: the series pair behaves as one spring of stiffness $k_{\text{eff}}$, so the triangle formula applies to it. $U=\frac{1}{2}(200)(0.09{)}^2=\frac{1}{2}(200)(0.0081)=0.81\text{J}$ Answer: $U=0.81\text{J}$.

What (a): the total mechanical energy of the oscillator equals the spring energy at maximum stretch, $x=A$ — there the block is momentarily still, so all energy sits in the spring as PE. Why (a): in Simple Harmonic Motion energy sloshes back and forth between spring PE and kinetic energy, but the total is constant; evaluating it at the turning point (where kinetic $=0$) is the easiest place to read the whole amount. $E=\frac{1}{2}k{A}^2=\frac{1}{2}(50)(0.12{)}^2=\frac{1}{2}(50)(0.0144)=0.36\text{J}$ So (a) $E=0.36\text{J}$. What (b): find the spring PE at $x=0.06\text{m}$, then subtract from the total to get the kinetic part, then divide. Why (b): at any position the energy splits into spring PE $=\frac{1}{2}k{x}^2$ and kinetic $=E-U$; the fraction kinetic is kinetic energy over total. $U=\frac{1}{2}(50)(0.06{)}^2=\frac{1}{2}(50)(0.0036)=0.09\text{J}$ $\text{kinetic}=E-U=0.36-0.09=0.27\text{J}$ $\text{fraction kinetic}=\frac{0.27}{0.36}=0.75=75\%$ So (b) $75\%$ of the energy is kinetic ($0.27\text{J}$ kinetic, $0.09\text{J}$ spring PE). Neat pattern: at half amplitude, PE is ${\left(\frac{1}{2}\right)}^2=\frac{1}{4}$ of the total, so kinetic is $\frac{3}{4}$. That $x^2$ dependence again!