SPH3U Unit 3: Energy and Society

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Comprehensive flashcards covering Work, Kinetic and Gravitational Potential Energy, Conservation of Mechanical Energy, Power, Energy Efficiency, and Thermal Energy Heat Exchange.

29 Terms

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How is energy defined in physics, and what is its standard SI unit?

Energy is the ability to do work, representing a force applied over a distance. The SI unit for energy is the Joule (J\text{J}), where 1 J=1 N⋅1 m1\,\text{J} = 1\,\text{N} \cdot 1\,\text{m}.

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What is the primary difference between potential energy and kinetic energy?

Potential energy is energy stored in an object due to its position relative to other objects, whereas kinetic energy is the working form of energy that an object possesses because of its motion.

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What formula is used to calculate the kinetic energy (EkE_k) of an object?

Ek=12mv2E_k = \frac{1}{2}mv^2, where mm is mass in kilograms (kg\text{kg}) and vv is velocity in meters per second (m/s\text{m/s}).

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Why can electrical energy be classified as both potential energy and kinetic energy?

It is potential energy when electrons are stationary in an open circuit waiting to move from the electrical panel to the outlet, and kinetic energy when a device is plugged in, closing the circuit and allowing electrons to move.

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What are the seven types of electromagnetic radiation that carry radiant energy, ranked from lowest to highest energy?

From lowest to highest energy: Radio Waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, and Gamma rays.

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What three conditions must be met for mechanical work (WW) to be done on an object?

A force must be applied, the object must undergo displacement, and a component of the applied force must be in the direction of motion (W=Fcos⁡(θ)ΔdW = F\cos(\theta)\Delta d).

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A tourist pulls luggage with a force of 55 N55\,\text{N} at an angle of 55∘55^\circ from the horizontal over a displacement of 125 m [E]125\,\text{m}\text{ [E]}. How much work is performed?

W=Fcos⁡(θ)Δd=(55 N)cos⁡(55∘)(125 m)=3.9×103 JW = F\cos(\theta)\Delta d = (55\,\text{N})\cos(55^\circ)(125\,\text{m}) = 3.9 \times 10^3\,\text{J}.

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A warehouse worker lifts an 11.6 kg11.6\,\text{kg} box vertically 97.5 cm97.5\,\text{cm}, carries it horizontally 11.2 m11.2\,\text{m} at constant speed, and places it back onto the floor. How much work is done to lift the box, and what is the total work done overall?

Work done to lift the box is W=mgΔd=(11.6 kg)(9.81 m/s2)(0.975 m)=111 JW = mg\Delta d = (11.6\,\text{kg})(9.81\,\text{m/s}^2)(0.975\,\text{m}) = 111\,\text{J}. Total work is 0 J0\,\text{J} because carrying it horizontally applies force perpendicular to motion (cos⁡(90∘)=0\cos(90^\circ) = 0), and lowering it back down does negative work of equal magnitude.

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A worker pushes two stacked boxes with a combined mass of 26 kg26\,\text{kg} across a concrete floor (μk=0.26\mu_k = 0.26) at a constant velocity for 4.5 m4.5\,\text{m}. How much work is done?

W=FappΔd=μkmgΔd=(0.26)(26 kg)(9.81 m/s2)(4.5 m)=3.0×102 JW = F_{\text{app}}\Delta d = \mu_k mg\Delta d = (0.26)(26\,\text{kg})(9.81\,\text{m/s}^2)(4.5\,\text{m}) = 3.0 \times 10^2\,\text{J}.

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A curler applies 22.0 N22.0\,\text{N} of force to a curling stone, accelerating it from rest to 3.10 m/s3.10\,\text{m/s} in 2.10 s2.10\,\text{s}. How much work does the curler do on the stone?

Displacement is Δd=(vf+vi2)Δt=(3.10 m/s+02)(2.10 s)=3.26 m\Delta d = \left(\frac{v_f + v_i}{2}\right)\Delta t = \left(\frac{3.10\,\text{m/s} + 0}{2}\right)(2.10\,\text{s}) = 3.26\,\text{m}. Work done is W=FΔd=(22.0 N)(3.26 m)=71.7 JW = F\Delta d = (22.0\,\text{N})(3.26\,\text{m}) = 71.7\,\text{J}.

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How is total mechanical energy (EmE_m) defined mathematically?

Mechanical energy is the sum of an object's kinetic energy and its gravitational potential energy: Em=Ek+EgE_m = E_k + E_g.

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An asteroid with a mass of 7.34×103 kg7.34 \times 10^3\,\text{kg} travels at 3.59×103 m/s3.59 \times 10^3\,\text{m/s}. How much kinetic energy does it possess, and how many sticks of dynamite (1.0×106 J1.0 \times 10^6\,\text{J} each) possess equivalent energy?

The asteroid possesses Ek=12(7.34×103 kg)(3.59×103 m/s)2=4.73×1010 JE_k = \frac{1}{2}(7.34 \times 10^3\,\text{kg})(3.59 \times 10^3\,\text{m/s})^2 = 4.73 \times 10^{10}\,\text{J}. This equals 4.73×1044.73 \times 10^4 sticks of dynamite.

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A .308 Winchester fires an 11.66 g11.66\,\text{g} bullet at 817.8 m/s817.8\,\text{m/s}, while a .300 Winchester Magnum fires the same bullet at 893.1 m/s893.1\,\text{m/s} through a 0.559 m0.559\,\text{m} barrel. What percentage greater recoil force does the .300 Winchester Magnum produce?

The .308 generates a recoil force of 6.97×103 N6.97 \times 10^3\,\text{N} (Ek=3.89×103 JE_k = 3.89 \times 10^3\,\text{J}), while the .300 Magnum generates 8.32×103 N8.32 \times 10^3\,\text{N} (Ek=4.65×103 JE_k = 4.65 \times 10^3\,\text{J}). The shooter experiences 19.3%19.3\% greater recoil force from the .300 Winchester Magnum.

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<p>Based on the diagram, a $$2250\,\text{kg}$$ roller coaster car starts from rest at Point A ($$30\,\text{m}$$). Assuming mechanical energy is conserved, what is its maximum speed at ground level?</p>

Based on the diagram, a 2250 kg2250\,\text{kg} roller coaster car starts from rest at Point A (30 m30\,\text{m}). Assuming mechanical energy is conserved, what is its maximum speed at ground level?

At Point A, total mechanical energy is Eg=mgh=(2250 kg)(9.81 m/s2)(30 m)=662175 JE_g = mgh = (2250\,\text{kg})(9.81\,\text{m/s}^2)(30\,\text{m}) = 662175\,\text{J}. At the ground, all potential energy becomes kinetic energy: v=2gh=2(9.81 m/s2)(30 m)=24.3 m/sv = \sqrt{2gh} = \sqrt{2(9.81\,\text{m/s}^2)(30\,\text{m})} = 24.3\,\text{m/s}.

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What does the Law of Conservation of Energy state regarding energy changes in an isolated system and its surroundings?

Energy cannot be created or destroyed; it can only change from one form to another (ΔE=0\Delta E = 0). Within any physical system, ΔEsystem=−ΔEsurroundings\Delta E_{\text{system}} = -\Delta E_{\text{surroundings}}.

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Two high jumpers, one weighing 70 kg70\,\text{kg} and the other 68 kg68\,\text{kg}, leave the ground with an initial vertical velocity of 4.8 m/s4.8\,\text{m/s}. Which jumper reaches a greater height?

Both jumpers reach the exact same height of 1.17 m1.17\,\text{m}. Setting 12mv2=mgh\frac{1}{2}mv^2 = mgh cancels out mass, giving h=v22g=(4.8 m/s)22(9.81 m/s2)=1.17 mh = \frac{v^2}{2g} = \frac{(4.8\,\text{m/s})^2}{2(9.81\,\text{m/s}^2)} = 1.17\,\text{m}.

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A 4.15 kg4.15\,\text{kg} IKEA Tarno folding chair is dropped from a balcony 92.4 m92.4\,\text{m} above the street. What maximum speed does it reach right before impact?

Using conservation of energy, v=2gh=2(9.81 m/s2)(92.4 m)=42.6 m/sv = \sqrt{2gh} = \sqrt{2(9.81\,\text{m/s}^2)(92.4\,\text{m})} = 42.6\,\text{m/s}.

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A hiker standing on a 3.7 m3.7\,\text{m} cliff throws a rock upward at 12 m/s12\,\text{m/s}. What is the rock's maximum height above the release point, and what is its maximum speed upon hitting the ground?

The rock reaches a maximum height of h=v22g=(12 m/s)22(9.81 m/s2)=7.3 mh = \frac{v^2}{2g} = \frac{(12\,\text{m/s})^2}{2(9.81\,\text{m/s}^2)} = 7.3\,\text{m} above the release point. Dropping from a total height of 3.7 m+7.3 m=11.0 m3.7\,\text{m} + 7.3\,\text{m} = 11.0\,\text{m}, its impact speed is v=2g(htotal)=15 m/sv = \sqrt{2g(h_{\text{total}})} = 15\,\text{m/s}.

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How is power (PP) defined in physics, and what units are used to measure it?

Power is a scalar quantity describing the rate at which energy transfer or work occurs: P=ΔEΔt=WΔtP = \frac{\Delta E}{\Delta t} = \frac{W}{\Delta t}. It is measured in Joules per second (J/s\text{J/s}), commonly called Watts (1 W=1 J/s1\,\text{W} = 1\,\text{J/s}).

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A powerlifter raises an 88.5 kg88.5\,\text{kg} barbell a vertical distance of 2.10 m2.10\,\text{m} in 0.94 s0.94\,\text{s}. How much power did the lifter exert?

P=mghΔt=(88.5 kg)(9.81 m/s2)(2.10 m)0.94 s=1.9×103 WP = \frac{mgh}{\Delta t} = \frac{(88.5\,\text{kg})(9.81\,\text{m/s}^2)(2.10\,\text{m})}{0.94\,\text{s}} = 1.9 \times 10^3\,\text{W}.

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A 60 W60\,\text{W} desk light bulb consumes 1.84×106 J1.84 \times 10^6\,\text{J} of energy before being turned off. How long was the bulb left on?

Δt=ΔEP=1.84×106 J60 W≈30667 s\Delta t = \frac{\Delta E}{P} = \frac{1.84 \times 10^6\,\text{J}}{60\,\text{W}} \approx 30667\,\text{s}, which equals 8.5 h8.5\,\text{h}.

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How is percent efficiency defined, and what is the efficiency of a 60 W60\,\text{W} incandescent bulb that loses 54 J54\,\text{J} to thermal energy and 0.8 J0.8\,\text{J} to sound energy each second?

% Efficiency=Euseful outEin×100%\%\,\text{Efficiency} = \frac{E_{\text{useful out}}}{E_{\text{in}}} \times 100\%. The useful light energy output is 60 J−54 J−0.8 J=5.2 J60\,\text{J} - 54\,\text{J} - 0.8\,\text{J} = 5.2\,\text{J}, resulting in an efficiency of 5.2 J60 J×100%=8.7%\frac{5.2\,\text{J}}{60\,\text{J}} \times 100\% = 8.7\%.

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A 78 kg78\,\text{kg} skateboarder rolls down a 6.2 m6.2\,\text{m} ramp and reaches a speed of 5.1 m/s5.1\,\text{m/s} at the bottom. What is the skateboarder's efficiency?

Input gravitational potential energy is Eg=mgh=(78 kg)(9.81 m/s2)(6.2 m)=4744 JE_g = mgh = (78\,\text{kg})(9.81\,\text{m/s}^2)(6.2\,\text{m}) = 4744\,\text{J}. Output kinetic energy is Ek=12(78 kg)(5.1 m/s)2=1014 JE_k = \frac{1}{2}(78\,\text{kg})(5.1\,\text{m/s})^2 = 1014\,\text{J}. The efficiency is 1014 J4744 J×100%=21%\frac{1014\,\text{J}}{4744\,\text{J}} \times 100\% = 21\%.

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What distinguishes temperature, thermal energy, and heat?

Temperature measures the average kinetic energy of the particles in a substance. Thermal energy is the sum of all kinetic and potential energies possessed by those particles. Heat is the transfer, release, or absorption of thermal energy between substances.

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What physical phenomena serve as the reference points for the Celsius, Fahrenheit, and Kelvin temperature scales?

Celsius is based on the freezing (0∘C0^\circ\text{C}) and boiling (100∘C100^\circ\text{C}) points of water; Fahrenheit is based on an equal ice:salt mixture (0∘F0^\circ\text{F}); and Kelvin is based on the total thermal energy possessed by a substance, where 0 K0\,\text{K} represents zero particle motion.

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What are the three mechanisms of heat exchange, and how do they function?

Conduction is direct energy transfer via colliding particles in physical contact. Convection is energy transfer via the circulating movement of fluid particles (liquids or gases) caused by density differences. Radiation is energy transfer via electromagnetic waves without requiring a medium.

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What is specific heat capacity (cc), and why does beach sand heat up and cool down much faster than ocean water?

Specific heat capacity is the quantity of heat needed to change the temperature of a specific mass by a set amount (Q=mcΔTQ = mc\Delta T). Beach sand has a low specific heat capacity (c=800 J/(kg⋅∘C)c = 800\,\text{J}/(\text{kg}\cdot^\circ\text{C})), requiring roughly one-fifth the energy of water (c=4186 J/(kg⋅∘C)c = 4186\,\text{J}/(\text{kg}\cdot^\circ\text{C})) to change temperature.

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A circuit breaker trips when 3.0 kg3.0\,\text{kg} of copper wire (c=0.385 J/(g⋅∘C)c = 0.385\,\text{J}/(\text{g}\cdot^\circ\text{C})) heats from 20∘C20^\circ\text{C} to 60∘C60^\circ\text{C}. How much excess electrical energy was absorbed by the wire?

Q=mcΔT=(3000 g)(0.385 J/(g⋅∘C))(60∘C−20∘C)=46200 J=4.62 kJQ = mc\Delta T = (3000\,\text{g})(0.385\,\text{J}/(\text{g}\cdot^\circ\text{C}))(60^\circ\text{C} - 20^\circ\text{C}) = 46200\,\text{J} = 4.62\,\text{kJ}.

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A 4.0 kg4.0\,\text{kg} metal bar heated to 100∘C100^\circ\text{C} is placed into a 100 g100\,\text{g} aluminum calorimeter (c=920 J/(kg⋅∘C)c = 920\,\text{J}/(\text{kg}\cdot^\circ\text{C})) with 500 g500\,\text{g} of water (c=4186 J/(kg⋅∘C)c = 4186\,\text{J}/(\text{kg}\cdot^\circ\text{C})) at 20∘C20^\circ\text{C}. If the final equilibrium temperature is 35∘C35^\circ\text{C}, is the bar made of pure gold (c=129 J/(kg⋅∘C)c = 129\,\text{J}/(\text{kg}\cdot^\circ\text{C}))?

Yes. Heat gained by the water and container is Qgained=(0.5 kg)(4186)(15∘C)+(0.1 kg)(920)(15∘C)=32775 JQ_{\text{gained}} = (0.5\,\text{kg})(4186)(15^\circ\text{C}) + (0.1\,\text{kg})(920)(15^\circ\text{C}) = 32775\,\text{J}. Setting Qlost=−QgainedQ_{\text{lost}} = -Q_{\text{gained}} yields (4.0 kg)c(65∘C)=32775 J(4.0\,\text{kg})c(65^\circ\text{C}) = 32775\,\text{J}, which gives c≈126 J/(kg⋅∘C)c \approx 126\,\text{J}/(\text{kg}\cdot^\circ\text{C}), confirming it is authentic gold or very close to pure.