A.3 Work, Power and Energy

Solution

1. Work done by the constant horizontal force:
2. $W = F s \cos\theta = 150 \times 4.0 \times \cos 0^{\circ} = 600\ \mathrm{J}$
3. With no other horizontal forces, the net work equals the change in kinetic energy:
4. $W_{\text{net}} = \Delta E_{\text{k}} = 600\ \mathrm{J}$
5. So the crate’s kinetic energy increases by
6. $600\ \mathrm{J}.$

Solution

1. Initial kinetic energy:
2. $E_{\text{k,i}} = \tfrac{1}{2}mv^{2} = \tfrac{1}{2}\times0.40\times8.0^{2} = 12.8\ \mathrm{J}$
3. At maximum height, speed is zero so all mechanical energy is gravitational potential:
4. $E_{\text{p,grav}} = mgh = 12.8\ \mathrm{J}$
5. Solve for h:
6. $h = \dfrac{E_{\text{p,grav}}}{mg} = \dfrac{12.8}{0.40\times9.8} \approx 3.3\ \mathrm{m}.$

Solution

1. (a) Elastic potential energy:
2. $E_{\text{p,elastic}} = \tfrac{1}{2}k(\Delta x)^{2} = \tfrac{1}{2}\times250\times0.060^{2} = 0.45\ \mathrm{J}$
3. (b) When the spring returns to natural length, all elastic energy becomes kinetic:
4. $E_{\text{k}} = 0.45\ \mathrm{J} = \tfrac{1}{2}mv^{2}$
5. Solve for v:
6. $v = \sqrt{\dfrac{2E_{\text{k}}}{m}} = \sqrt{\dfrac{2\times0.45}{0.50}} \approx 1.34\ \mathrm{m\,s^{-1}}.$

Solution

1. (a) Work done against gravity:
2. $W = mgh = 900\times9.8\times12 = 1.06\times10^{5}\ \mathrm{J}\ \text{(3 s.f.)}$
3. (b) Power as rate of doing work:
4. $P = \dfrac{W}{t} = \dfrac{1.06\times10^{5}}{18} \approx 5.9\times10^{3}\ \mathrm{W} = 5.9\ \mathrm{kW}.$

Solution

1. (a) Useful power is the rate of gaining gravitational potential energy:
2. $P_{\text{useful}} = Fv = mgv = 40\times9.8\times1.2 \approx 470\ \mathrm{W}$
3. (b) Efficiency in terms of power:
4. $\eta = \dfrac{P_{\text{useful}}}{P_{\text{input}}} = \dfrac{470}{750} \approx 0.63 = 63\%.$

Solution

1. Electrical output energy in one hour:
2. $E_{\text{out}} = P_{\text{out}} t = 800\times3600 = 2.88\times10^{6}\ \mathrm{J}$
3. Efficiency links output and input energy:
4. $E_{\text{out}} = \eta E_{\text{in}} \Rightarrow E_{\text{in}} = \dfrac{2.88\times10^{6}}{0.25} = 1.15\times10^{7}\ \mathrm{J}$
5. Energy density is energy per unit volume, so volume of fuel used is:
6. $V = \dfrac{E_{\text{in}}}{\text{energy density}} = \dfrac{1.15\times10^{7}}{3.2\times10^{7}} \approx 0.36\ \mathrm{m^{3}}.$