Energy Changes in a System, and The Ways Energy is Stored Before and After such Changes

Solution

1. 1

   Identify the initial store before the ball is released.

   $\text{Initial: gravitational PE store of the ball–Earth system is full.}$

2. 2

   Describe the fall.

   $\text{During the fall: gravitational PE empties; kinetic store of the ball fills.}$

3. 3

   Describe the bounce.

   $\text{At the floor: kinetic store empties; elastic PE store of the deforming ball briefly fills; then re-empties back to kinetic.}$

4. 4

   Account for why the bounce is shorter.

   $\text{Each bounce: some energy is transferred to thermal stores of the ball and floor (friction inside the rubber), and by sound radiation. So less energy returns to gravitational PE on each rebound.}$

Solution

1. 1

   Input store and pathway.

   $\text{Energy enters the fan electrically (transfer pathway).}$

2. 2

   Useful output store.

   $\text{Useful: kinetic store of the fan blades (and the air the blades push).}$

3. 3

   Wasted output stores.

   $\text{Wasted: thermal store of the motor (resistive heating), thermal store of the air (eventually), sound radiation.}$

Solution

1. 1

   Kinetic store at terminal velocity.

   $\text{At terminal velocity, the speed is constant, so the kinetic store of the skydiver is NOT changing.}$

2. 2

   Apply conservation.

   $\text{All of the gravitational PE leaving the system must be going somewhere else.}$

3. 3

   Identify the destination.

   $\text{Air resistance (a force) does negative work on the skydiver, transferring energy to the thermal store of the air around her by heating (collisions with air molecules).}$

Solution

1. 1

   Write the formula.

   $E_k = \tfrac{1}{2} m v^2$

2. 2

   Substitute.

   $E_k = 0.5 \times 0.020 \times 350^2$

3. 3

   Square first, then multiply.

   $E_k = 0.5 \times 0.020 \times 122{,}500 = 1225\,\text{J}$

Solution

1. 1

   Apply conservation of energy.

   $mgh = \tfrac{1}{2}mv^2$

2. 2

   Cancel m and rearrange.

   $v = \sqrt{2gh}$

3. 3

   Substitute g = 9.8 N/kg, h = 30 m.

   $v = \sqrt{2 \times 9.8 \times 30} = \sqrt{588} \approx 24.2\,\text{m/s}$

Solution

1. 1

   Convert extension to metres.

   $e = 8\,\text{cm} = 0.08\,\text{m}$

2. 2

   Apply the formula.

   $E_e = \tfrac{1}{2} \times 120 \times 0.08^2$

3. 3

   Square the extension; multiply.

   $E_e = 60 \times 0.0064 = 0.384\,\text{J}$

Solution

1. 1

   Recall the kinetic energy formula.

   $E_k = \tfrac{1}{2} m v^2$

2. 2

   Substitute m = 75 kg and v = 6 m/s.

   $E_k = \tfrac{1}{2} \times 75 \times 6^2$

3. 3

   Evaluate (square the speed first, then multiply).

   $E_k = 37.5 \times 36 = 1350\,\text{J}$

Solution

1. 1

   Recall the formula.

   $E_p = m g h$

2. 2

   Substitute m = 1.2 kg, g = 9.8 N/kg, h = 2.4 m.

   $E_p = 1.2 \times 9.8 \times 2.4$

3. 3

   Multiply.

   $E_p = 28.224\,\text{J}$

Solution

1. 1

   Convert the extension to metres.

   $e = 10\,\text{cm} = 0.10\,\text{m}$

2. 2

   Use the elastic PE formula (on the AQA equation sheet).

   $E_e = \tfrac{1}{2} k e^2$

3. 3

   Substitute and evaluate (remember to square the extension).

   $E_e = \tfrac{1}{2} \times 50 \times (0.10)^2 = 25 \times 0.01 = 0.25\,\text{J}$

Solution

1. 1

   Convert mass to kilograms.

   $m = 250\,\text{g} = 0.250\,\text{kg}$

2. 2

   Calculate the temperature change Δθ.

   $\Delta\theta = 95 - 18 = 77\,\text{°C}$

3. 3

   Apply ΔE = mcΔθ.

   $\Delta E = m c \Delta\theta = 0.250 \times 4200 \times 77$

4. 4

   Evaluate.

   $\Delta E = 80{,}850\,\text{J} \approx 80.9\,\text{kJ}$

Solution

1. 1

   Find the energy needed to heat the water.

   $\Delta E = 0.50 \times 4200 \times 80 = 168{,}000\,\text{J}$

2. 2

   Convert power to joules per second.

   $P = 2.4\,\text{kW} = 2400\,\text{W} = 2400\,\text{J/s}$

3. 3

   Time = energy ÷ power.

   $t = \dfrac{\Delta E}{P} = \dfrac{168{,}000}{2400} = 70\,\text{s}$

Solution

1. 1

   Convert time to seconds.

   $t = 2 \times 3600 = 7200\,\text{s}$

2. 2

   Apply E = Pt.

   $E = 40 \times 7200 = 288{,}000\,\text{J} = 288\,\text{kJ}$

Solution

1. 1

   Find the work done (= gravitational PE gained).

   $W = mgh = 250 \times 9.8 \times 6 = 14{,}700\,\text{J}$

2. 2

   Apply P = W/t.

   $P = 14{,}700 / 30 = 490\,\text{W}$

Solution

1. 1

   Convert to base units.

   $P = 2000\,\text{W};\quad E = 500{,}000\,\text{J}$

2. 2

   Apply t = E/P.

   $t = 500{,}000 / 2000 = 250\,\text{s} \approx 4\,\text{min }10\,\text{s}$