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Work is the result of exerting a force to move an object in the direction of the force
\[Work = Force \times displacement \qquad W=Fd\]The definition of energy is actually very complicated and not easy to understand
In short, Energy is a use of forces to accomplish some task, and work is one way of transferring energy from one object to another.
Types of energy:
Kinetic Energy: The energy of motion
\[KE = \tfrac{1}{2}mv^2\]PE in gravity
\[PE = F_g d = (mg)h = mgh\]Mechanical Energy is the combination of energy resulting in motion ($ME = KE + PE$)
Generally, we deal with an interchange (conversion) of energy between KE and PE
\[W = \Delta KE\]Demo: Roll a ball - exert work to gain energy
However, there are other forces that could be involved. For example, if friction is involved, then this interaction causes a loss of heat to thermal energy:
\[W = \Delta (KE+PE+TE)\]What is the kinetic energy of a 1500-kg car traveling at 27 m/s (about 60 mph)?
\[\begin{align*}KE &= \tfrac{1}{2} 1500(27^2) \\ &= 546,750 J \\ &= 546.75 kJ\end{align*}\]
What is the total mechanical energy of a 12,000-kg airplane traveling at 250 m/s (about 560 mph)at an altitude of 16 km (about 10 miles)?
\[\begin{align*}ME &= KE + PE \\ &= \tfrac{1}{2} mv^2 + mgh \\ &= \tfrac{1}{2} 12000(250^2) + 12000(9.8)(16) \\ &= 375,000,000 J + 1,881,600 J \\ &= 376,881,600 J \\ &= 376.9 MJ\end{align*}\]
A 40-kg child is sitting on a 30-kg pushcart. What force is needed to bring the child and pushcart up to a speed of 10 m/s?
\[\begin{align*}W &= \Delta (KE + PE) \\ &= (KE_f + PE_f) - (KE_0 + KE_0) \\ &= KE_f - KE_0 \\ &= \tfrac{1}{2} mv_f^2 - \tfrac{1}{2} mv_i^2 \\ &= \tfrac{1}{2} 70(10^2 ) - \tfrac{1}{2} 70(0^2 ) \\ &= \tfrac{1}{2} 70(100) \\ &= 3,500 J\end{align*}\]
Applying forces isn’t the only way to transfer energy. Heat (indicated by Q) is energy that is transferred due to a difference in temperatures. So, this also plays a role in our transfer of energy:
\[W + Q = \Delta (KE+PE+TE)\]More on heat will be addressed in future lectures.
Power is the rate at which energy is transferred
\[Power = \frac{work~done}{time~taken} = \frac{energy~used}{time~taken}\]The main unit for power is the Watt:
\[1W = \frac{1 J}{1 sec}\]1 Watt is the equivalent of lifting one apple by one meter every second.
I lift a box of 20 apples (0.1 kg each) from the floor to a shelf 2.5 m high in a time of 4.0 sec. How much work have I done?
\[\begin{align*}W &= \Delta (KE+PE) \\ &= PE_f - PE_i \\ &= mgh_f - mgh_i \\ &= (20*0.1kg)(10)2.5m - 0 \\ &= 50J\end{align*}\]What was my power output?
\[Power = \frac{work~done}{time} = \frac{50 J}{4 s} = 12.5 W\]
We can use the power output of some device and the time used to find out the energy used to run the device:
\[Energy~used~=~Power \times time\]This produces a common unit of energy when calculating it this way: the kilowatt-hour (kWh).
An electric heater has a power rating of 1500 W (1.5kW). If electricity costs $0.12 per kWh, how much will it cost to run the heater for 6 hours?
\[energy = 1500 W \times 6 hrs = 1.5 kW \times 6 hrs = 9.0 kWh\] \[9.0 kWh\times \frac{\$0.12}{1 kWh}=\$1.08\]
Energy per capita and power output per capita are the average energy/power output per person.
What is the average energy consumption and power output of people in the U.S.? (per capita energy consumption and per capita power output)
\[energy = 97.8~\tfrac{quads}{year} \times \frac{1055*10^{15}~J}{1~quads} = 1.03*10^{20}\tfrac{J}{year}\] \[\begin{align*}energy~per~capita &= \frac{1.03*10^{20}~J/year}{332,000,000~people} \\ &= 3.1078*10^{11} \tfrac{J/person}{year} \\&= 310.78 \tfrac{GJ/person}{year}\end{align*}\] \[\begin{align*}power~output &= \frac{energy~used}{time} \\&= \frac{3.1078*10^{11}J/person}{1 year \times \tfrac{365 days}{1 year} \times \tfrac{24 hours}{1 day} \times \frac{3600 s}{1 hour}} \\&= 9855~W/person\end{align*}\]
- U.S. population: 332 million
- Total energy usage in one year: 97.8 quads