Spring Potential Energy — Practice Problem Solutions

1. $U_s=\tfrac12(400)(0.050)^2=0.50\text{ J}$. $U_s=0.50\text{ J}$.

2. $U_s\propto x^2$; tripling $x$ multiplies $U_s$ by $9$: $9\times2.0=18\text{ J}$. $18\text{ J}$.

3. Equal. $U_s=\tfrac12kx^2$ depends on the square of the displacement, so stretching and compressing by the same amount store the same energy ($\tfrac12k(0.10)^2$ either way).

4. $U_s(0.10)=\tfrac12(100)(0.10)^2=0.50\text{ J}$; $U_s(0.20)=\tfrac12(100)(0.20)^2=2.0\text{ J}$. $\Delta U_1=0.50\text{ J}$, $\Delta U_2=1.5\text{ J}$ — the second step is three times the first ($\Delta U_2-\Delta U_1=1.0\text{ J}=kd^2$).

5. $U_s(\pm0.30)=\tfrac12(80)(0.30)^2=3.6\text{ J}$; $U_s(0)=0$. $3.6\text{ J}\to0\to3.6\text{ J}$ — decreases by $3.6\text{ J}$, then increases by $3.6\text{ J}$; net change zero.

6. $x=mg/k=(0.20)(10)/40=0.050\text{ m}$. $\Delta U_g=-mgx=-(0.20)(10)(0.050)=-0.10\text{ J}$. $\Delta U_s=\tfrac12(40)(0.050)^2=+0.050\text{ J}$. $x=0.050\text{ m}$, $\Delta U_g=-0.10\text{ J}$, $\Delta U_s=+0.050\text{ J}$.

7. a) $F$ rises linearly from $0$ to $(50)(0.40)=20\text{ N}$; area $=\tfrac12(0.40)(20)=4.0\text{ J}$. b) $\tfrac12kx^2=\tfrac12(50)(0.40)^2=4.0\text{ J}$ ✓. c) Force at $0.20\text{ m}$ is $10\text{ N}$; the region is a trapezoid: $\tfrac12(10+20)(0.20)=3.0\text{ J}$. $3.0\text{ J}$ (check: $\tfrac12(50)(0.16-0.04)=3.0\text{ J}$).

8. a) $\Delta U_1=\tfrac12k\big[(x_0+d)^2-x_0^2\big]=\tfrac12k(2x_0d+d^2)$; $\Delta U_2=\tfrac12k\big[(x_0+2d)^2-(x_0+d)^2\big]=\tfrac12k(2x_0d+3d^2)$. b) $\Delta U_2-\Delta U_1=\tfrac12k(2d^2)=kd^2$. c) $kd^2$ contains no $x_0$ and is always positive, so $\Delta U_2>\Delta U_1$ for any original compression — the comparison can be made without knowing $x_0$.

9. The person does positive work on the block/spring, while the spring force does an equal amount of negative work; the net work is zero, so $\Delta K=0$ (work-energy theorem). The energy the person supplied doesn't become kinetic — it is stored as the spring's potential energy, $\Delta U_s=W_{person}$.

10. (C) $U_s=\tfrac12kx^2$ is zero at the relaxed length, $x=0$, and positive for compression and stretch alike; moving from $-A$ to $0$ it decreases to zero, and moving from $0$ to $+A$ it increases again.

11. (C) $\Delta U_g<0$ (the egg moves lower) and $\Delta U_s>0$ (the spring goes from relaxed to compressed). Both potential energies change, in opposite directions.