Conservation of Mechanical Energy — Practice Problem Solutions

1. System: ball–Earth. $mgh=\tfrac12mv^2\Rightarrow v=\sqrt{2gh}=\sqrt{2(10)(20)}=\sqrt{400}=20\text{ m/s}$. $v=20\text{ m/s}$.

2. a) $v=\sqrt{2gh}$. b) No to both. The mass cancels, and the ramp's angle never appears in the energy equation — only the height matters.

3. System: sled–Earth. $v_f=\sqrt{v_i^2-2gh}=\sqrt{36-2(10)(1.0)}=\sqrt{16}=4.0\text{ m/s}$. $v_f=4.0\text{ m/s}$.

4. System: block–spring. $U_s=\tfrac12(100)(0.20)^2=2.0\text{ J}=\tfrac12mv^2\Rightarrow v=\sqrt{2(2.0)/0.20}=\sqrt{20}\approx4.5\text{ m/s}$. $v\approx4.5\text{ m/s}$.

5. System: block–spring–Earth. $U_s=\tfrac12(400)(0.10)^2=2.0\text{ J}=mgh\Rightarrow h=\dfrac{2.0}{(0.40)(10)}=0.50\text{ m}$. $h=0.50\text{ m}$.

6. $E=mgh_A=(2.0)(10)(5.0)=100\text{ J}$. At C: $U=(2.0)(10)(3.0)=60\text{ J}$, so $K_C=40\text{ J}$, $v_C=\sqrt{2(40)/2.0}=\sqrt{40}\approx6.3\text{ m/s}$. At B: $K_B=100\text{ J}$, $v_B=\sqrt{2(100)/2.0}=10\text{ m/s}$. $E=100\text{ J}$, $K_C=40\text{ J}$, $v_C\approx6.3\text{ m/s}$, $v_B=10\text{ m/s}$.

7. a) Gravitational potential energy and spring potential energy (no kinetic energy — it's at rest). b) Only spring potential energy. c) Earth is not part of the block–spring system, so gravitational potential energy isn't part of that system's energy; gravity is just an external force acting on it. Which energies are on your list depends entirely on what you define as the system.

8. a) $\Delta E_{tot}=0$ — no external forces act on the asteroid–planet system and no energy is converted to internal energy. b) $\Delta U_g<0$ (closer means $r$ decreases and $U=-Gm_1m_2/r$ becomes more negative), and since $\Delta K+\Delta U_g=0$, $\Delta K=-\Delta U_g>0$: the asteroid speeds up as the system's potential energy decreases.

9. The claim is wrong. $\tfrac12mv_i^2=\tfrac12mv_f^2+mgh$ — $m$ appears in every term and cancels, giving $v_f=\sqrt{v_i^2-2gh}$ regardless of mass. A heavier sled has more kinetic energy and needs more energy to climb, and these exactly balance.

10. (B) System: sled–Earth, no friction, so $\tfrac12mv_i^2=\tfrac12mv_f^2+mgh\Rightarrow v_f=\sqrt{v_i^2-2gh}$ (it gains height, so it slows).

11. (B) $E_1=E_0$. For the block–spring–Earth system, there are no external forces doing work and friction is negligible, so total mechanical energy is unchanged; the energy just changes form (spring and gravitational potential to kinetic).