Nonconservative Forces and Energy Transfer — Practice Problem Solutions

1. $E_i=mgh=(2.0)(10)(1.5)=30\text{ J}$. $E_f=\tfrac12mv^2=\tfrac12(2.0)(16)=16\text{ J}$. $E_{dis}=30-16=14\text{ J}$. $E_i=30\text{ J}$, $K_f=16\text{ J}$, $E_{dis}=14\text{ J}$.

2. $E_{dis}=f,d\Rightarrow f=14/3.0\approx4.7\text{ N}$. $f\approx4.7\text{ N}$.

3. $E_i=\tfrac12(0.20)(25)=2.5\text{ J}$; $E_f=mgh=(0.20)(10)(0.50)=1.0\text{ J}$; $E_{dis}=1.5\text{ J}$; $f=1.5/1.5=1.0\text{ N}$. $E_{dis}=1.5\text{ J}$, $f=1.0\text{ N}$.

4. $W_{person}=Fd=(40)(5.0)=200\text{ J}$. Constant speed: $\Delta K=0$; $\Delta U_g=0$ (horizontal). All $200\text{ J}$ becomes internal energy of the crate and floor (friction, external). $W_{person}=200\text{ J}$, $\Delta K=0$, energy $\to$ internal.

5. $\Delta K=0$; $\Delta U_g=(80)(10)(-50)=-40{,}000\text{ J}$; $\Delta E_{mech}=-40{,}000\text{ J}$. $\Delta E_{mech}=-40{,}000\text{ J}$.

6. a) $\Delta U_g=(3.0)(10)(2.0)=60\text{ J}$; $\Delta K=0$; $W_{person}=\Delta E_{sys}=60\text{ J}$. b) Person $+60\text{ J}$ and gravity $-60\text{ J}$: net work $=0$.

7. a) $\tfrac12mv_0^2-E_{dis}=mgh$. b) $E_{dis}=fd=\tfrac12mv_0^2-mgh=\dfrac{m(v_0^2-2gh)}{2}\Rightarrow f=\dfrac{m(v_0^2-2gh)}{2d}$. c) $f=\dfrac{0.30(16-2(10)(0.40))}{2(1.0)}=\dfrac{0.30(8)}{2}=1.2\text{ N}$. $f=1.2\text{ N}$.

8. a) System: block–Earth. External force doing work: friction from the ramp (yes). Internal-energy conversion: yes. Mechanical energy is not conserved — it decreases even though the block speeds up ($K$ rises, but $U_g$ falls by more). b) Yes — block–Earth–ramp includes the surfaces that warm up, so the mechanical energy lost appears as internal energy inside the system; total energy (mechanical + internal) is conserved.

9. $K$: constant (constant speed). $U_g$: decreases steadily (straight line down). $E_{mech}=K+U_g$: decreases steadily, parallel to $U_g$. The force that changes $E_{mech}$ is air drag (external), converting mechanical energy to internal energy of the air and parachute.

10. (A) Friction from the ramp is a force external to the block–Earth system that dissipates mechanical energy into internal energy. (C) and (D) confuse total and mechanical energy; (B) is true but is not a justification for a decrease, since $U_g$ decreasing while $K$ increases could still conserve $E_{mech}$.

11. (C) Mechanical energy is $K+U$: $K$ stays constant while $U$ decreases, so mechanical energy decreases — air drag converts it to internal energy.