David Sang and Graham Jones Solutions for Chapter: Work, Energy and Power, Exercise 11: EXAM-STYLE QUESTIONS

 Use the equations of motion to show that the kinetic energy of an object of mass $m$ moving with velocity $v$ is $\frac{1}{2}m{v}^2$.

 A car of mass $800 \mathrm{kg}$ accelerates from rest to a speed of $20{ \mathrm{ms}}^{-1}$ in a time of $6.0 \mathrm{s}$.Calculate the average power used to accelerate the car in the first $6.0 \mathrm{s}$

 A car of mass $800 \mathrm{kg}$ accelerates from rest to a speed of $20{ \mathrm{ms}}^{-1}$ in a time of $6.0 \mathrm{s}$.The power passed by the engine of the car to the wheels is constant. Explain why the acceleration of the car decreases as the car accelerates.

What is the definition of potential energy.

 Identify differences between gravitational potential energy and elastic potential energy.

Seawater is trapped behind a dam at high tide and then released through turbines. The level of the water trapped by the dam falls $10.0 \mathrm{m}$ until it is all at the same height as the sea. Calculate the mass of seawater covering an area of $1.4\times {10}^6{ \mathrm{m}}^2$ and with a depth of $10.0 \mathrm{m}$. (Density of seawater $=1030 \mathrm{kg}{ \mathrm{m}}^{-3}$.)

Seawater is trapped behind a dam at high tide and then released through turbines. The level of the water trapped by the dam falls $10.0 \mathrm{m}$ until it is all at the same height as the sea. Calculate the maximum loss of potential energy of the seawater when passed through the turbines and the level of water on both side becomes same mass of water transferred is $1.442\times {10}^{10}kg$ .

Seawater is trapped behind a dam at high tide and then released through turbines. The level of the water trapped by the dam falls $10.0 \mathrm{m}$ until it is all at the same height as the sea. The potential energy of the seawater, $1.442\times {10}^{12} J$, is lost over a period of $6.0$ hours. Estimate the average power output of the power station over this time period, given that the efficiency of the power station is $50\%$.