State the condition for the work done by a force on a moving object to be zero.
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The force acts perpendicular to the direction of motion / the displacement is zero.
Work Energy Power And Efficiency
A box is dragged 6.0 m across a floor by a rope at 25° above horizontal. The tension is 75 N. Calculate the work done by the tension.
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$$\Delta W=F\Delta scos\theta =75\times 6.0\times cos{25}^{∘}$$ $$\Delta W=408\text{ J}\approx 410\text{ J (2 s.f.)}$$
Explain why a satellite in a circular orbit has constant kinetic energy, even though a gravitational force acts on it continuously.
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The gravitational force acts towards the centre of the orbit / perpendicular to the velocity; cos 90° = 0 so no work is done on the satellite by the gravitational force; with no net work done, the kinetic energy and therefore the speed remain constant.
A car of mass 1200 kg accelerates from 12 m s⁻¹ to 24 m s⁻¹. State the factor by which its kinetic energy increases.
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Factor of 4 / it quadruples (because Eₖ ∝ v²).
Calculate the kinetic energy of a 0.058 kg tennis ball moving at 54 m s⁻¹.
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$${E}_{k}=\frac{1}{2}m{v}^{2}=\frac{1}{2}\times 0.058\times {54}^{2}$$ $${E}_{k}=84.6\text{ J}\approx 85\text{ J (2 s.f.)}$$
Define the change in gravitational potential energy of an object near the Earth’s surface.
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The work done against gravity when the object is raised through a vertical height / the energy transferred to the gravitational potential store.
A 0.45 kg book is lifted from the floor onto a shelf 2.1 m above. Calculate the gain in gravitational potential energy.
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$$\Delta {E}_{grav}=mg\Delta h=0.45\times 9.81\times 2.1$$ $$\Delta {E}_{grav}=9.27\text{ J}\approx 9.3\text{ J (2 s.f.)}$$
Show that an object falling freely from rest through a vertical height of 5.0 m reaches a speed of approximately 9.9 m s⁻¹.
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$$mg\Delta h=\frac{1}{2}m{v}^{2}$$ $$v=\sqrt{2g\Delta h}=\sqrt{2\times 9.81\times 5.0}$$ $$v=9.905{\text{ m s}}^{-1}\approx 9.9{\text{ m s}}^{-1}\text{ as required}$$
A 65 kg cyclist coasts down a hill that drops 12 m vertically, starting from rest. The cyclist reaches a speed of 12 m s⁻¹ at the bottom. Determine the energy transferred to thermal stores due to friction and air resistance.
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Finding the loss in gravitational potential energy $$\Delta {E}_{grav}=mg\Delta h=65\times 9.81\times 12$$ $$\Delta {E}_{grav}=7651\text{ J}$$ Finding the kinetic energy gained $${E}_{k}=\frac{1}{2}m{v}^{2}=\frac{1}{2}\times 65\times {12}^{2}$$ $${E}_{k}=4680\text{ J}$$ Finding the energy transferred to thermal stores $$\Delta {E}_{grav}-{E}_{k}=7651-4680=2971\text{ J}\approx 3.0\times {10}^{3}\text{ J (2 s.f.)}$$
State the SI unit of power and define it in terms of base SI units.
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Watt / W; equivalent to J s⁻¹ / kg m² s⁻³.
A weightlifter raises a 95 kg barbell through 1.8 m in 1.4 s. Calculate the average useful power developed.
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Finding the work done against gravity $$W=mg\Delta h=95\times 9.81\times 1.8$$ $$W=1677\text{ J}$$ Finding the average useful power $$P=\frac{W}{t}=\frac{1677}{1.4}$$ $$P=1198\text{ W}\approx 1.2\times {10}^{3}\text{ W (2 s.f.)}$$
Explain why no real engine can have an efficiency of 100%.
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Some energy is always transferred to non-useful stores / thermal energy due to friction or resistance; therefore the useful output is always less than the total input.
A motor lifts a 12 kg load through 3.5 m in 6.0 s, drawing 95 W of electrical power. Calculate its efficiency.
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Finding the useful work done $$mg\Delta h=12\times 9.81\times 3.5=412\text{ J}$$ Finding the useful power output $$\frac{412}{6.0}=68.7\text{ W}$$ Finding the efficiency $$\text{efficiency}=\frac{68.7}{95}=0.72\text{ / }72\%$$
Deduce whether a kettle rated at 2.2 kW can boil 0.50 kg of water (initially at 20 °C) within 90 s, if the kettle is 78% efficient. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
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Finding the energy required to heat the water $$mc\Delta T=0.50\times 4200\times 80=168000\text{ J}$$ Finding the useful power output $$0.78\times 2200=1716\text{ W}$$ Finding the time required $$\frac{168000}{1716}=97.9\text{ s}$$ Since 97.9 s > 90 s, the kettle cannot boil the water within 90 s.
A wind turbine experiences a wind of speed 11 m s⁻¹ flowing through its blades. The mass of air passing through the blades each second is 1.6 × 10⁴ kg. The turbine generates 850 kW of useful electrical power. In your answer you should: calculate the kinetic energy delivered to the blades each second; calculate the efficiency of the turbine; explain, with reference to energy conservation, why the efficiency cannot reach 100%.
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Finding the kinetic energy delivered each second (input power) $$\frac{1}{2}m{v}^{2}=\frac{1}{2}\times 1.6\times {10}^{4}\times {11}^{2}$$ $$=9.68\times {10}^{5}\text{ W}$$ Finding the efficiency $$\text{efficiency}=\frac{850\times {10}^{3}}{9.68\times {10}^{5}}=0.878\text{ / }88\%$$ The air leaving the blades must still have some kinetic energy / the air cannot be brought to a complete stop; if the air stopped, no further air could pass through the blades; some energy is also transferred to thermal stores by friction in the bearings and electrical resistance in the generator; therefore the total useful output is always less than the kinetic energy input.