Energy

Elastic (strain) store.

The internal (thermal) store decreases; the particles in the tea lose kinetic energy as the tea cools; therefore the total kinetic and potential energy of the particles inside the tea decreases.

Energy is transferred from the chemical store of the battery to the internal (thermal) store and the kinetic store of the buzzer by electrical currents (electrical work done); energy is also transferred from the buzzer to the surroundings by sound waves.

Energy is transferred from the kinetic store of the car to the internal (thermal) store of the brakes mechanically (by forces); the brake pads and discs increase in temperature because friction between surfaces causes particles to vibrate faster; therefore the internal (thermal) store of the brakes and surroundings increases.

Energy cannot be created or destroyed; it can only be transferred from one store to another.

Finding the wasted energy $${E}_{\text{wasted}}={E}_{\text{input}}-{E}_{\text{useful}}$$ Given: $${E}_{\text{input}}=50\text{ J}$$ $${E}_{\text{useful}}=5\text{ J}$$ Substituting: $${E}_{\text{wasted}}=50-5$$ $${E}_{\text{wasted}}=45\text{ J}$$ The wasted energy is 45 J; it is transferred to the internal (thermal) store of the surroundings by heating.

Converting speed to SI units: $$v=\frac{90}{3.6}$$ $$v=25{\text{ m s}}^{-1}$$ Finding the kinetic energy Equation used: $${E}_{k}=\frac{1}{2}m{v}^{2}$$ Given: $$m=1200\text{ kg}$$ $$v=25{\text{ m s}}^{-1}$$ Substituting: $${E}_{k}=\frac{1}{2}\times 1200\times (25{)}^{2}$$ $${E}_{k}=\frac{1}{2}\times 1200\times 625$$ $${E}_{k}=375 000\text{ J}$$ The kinetic energy is 375 000 J (375 kJ).

Finding the speed Equation used: $${E}_{k}=\frac{1}{2}m{v}^{2}$$ Rearranging for speed: $$v=\sqrt{\frac{2{E}_{k}}{m}}$$ Given: $${E}_{k}=4.0\text{ J}$$ $$m=0.50\text{ kg}$$ Substituting: $$v=\sqrt{\frac{2\times 4.0}{0.50}}$$ $$v=\sqrt{\frac{8.0}{0.50}}$$ $$v=\sqrt{16}$$ $$v=4.0{\text{ m s}}^{-1}$$ The speed of the ball is 4.0 m s⁻¹.

Converting mass to SI units: $$m=\frac{500}{1000}$$ $$m=0.50\text{ kg}$$ Finding the change in gravitational potential energy Equation used: $$\Delta {E}_{p}=mg\Delta h$$ Given: $$m=0.50\text{ kg}$$ $$g=9.8{\text{ N kg}}^{-1}$$ $$\Delta h=4.0\text{ m}$$ Substituting: $$\Delta {E}_{p}=0.50\times 9.8\times 4.0$$ $$\Delta {E}_{p}=19.6\text{ J}$$ $$\Delta {E}_{p}\approx 20\text{ J (2 s.f.)}$$ The gravitational potential energy lost is 20 J.

The minimum energy transferred equals the gain in gravitational potential energy (by conservation of energy, assuming no energy is wasted). Finding the change in gravitational potential energy Equation used: $$\Delta {E}_{p}=mg\Delta h$$ Given: $$m=250\text{ kg}$$ $$g=9.8{\text{ N kg}}^{-1}$$ $$\Delta h=12\text{ m}$$ Substituting: $$\Delta {E}_{p}=250\times 9.8\times 12$$ $$\Delta {E}_{p}=29 400\text{ J}$$ $$\Delta {E}_{p}\approx 29 000\text{ J (2 s.f.)}$$ The minimum energy transferred by the motor is 29 000 J (29 kJ).

Finding the wasted energy $${E}_{\text{wasted}}={E}_{\text{input}}-{E}_{\text{useful}}$$ $${E}_{\text{wasted}}=800-520$$ $${E}_{\text{wasted}}=280\text{ J}$$ The wasted energy is 280 J; it is transferred to the internal (thermal) store of the motor and surroundings by heating.

By conservation of energy, the gravitational potential energy lost equals the kinetic energy gained at the lowest point. Step 1: Finding the gravitational potential energy lost Equation used: $$\Delta {E}_{p}=mg\Delta h$$ Given: $$m=0.30\text{ kg}$$ $$g=9.8{\text{ N kg}}^{-1}$$ $$\Delta h=0.40\text{ m}$$ Substituting: $$\Delta {E}_{p}=0.30\times 9.8\times 0.40$$ $$\Delta {E}_{p}=1.176\text{ J}$$ This equals the kinetic energy at the lowest point: $${E}_{k}=1.176\text{ J}$$ Step 2: Finding the speed Equation used: $${E}_{k}=\frac{1}{2}m{v}^{2}$$ Rearranging for speed: $$v=\sqrt{\frac{2{E}_{k}}{m}}$$ Substituting: $$v=\sqrt{\frac{2\times 1.176}{0.30}}$$ $$v=\sqrt{\frac{2.352}{0.30}}$$ $$v=\sqrt{7.84}$$ $$v=2.8{\text{ m s}}^{-1}$$ The speed of the bob at the lowest point is 2.8 m s⁻¹.