17. Conservation of energy

From Mechanics

Revision as of 14:05, 30 September 2009

Key Results

Kinetic Energy \displaystyle \frac{1}{2}m{{v}^{2}}

Potential Energy mgh

Energy lost due to friction = Work Done Against Friction

Example 17.1

A ball, of mass 300 grams, is hit from ground level, so that its initial speed is 16 \displaystyle \text{m}{{\text{s}}^{-1}}. At its maximum height the ball travels at 6 \displaystyle \text{m}{{\text{s}}^{-1}}.

(a) Find the maximum and minimum kinetic energy of the ball.

(b) If the ball had been projected vertically upwards what would be the minimum kinetic energy of the ball?

Solution (a) The kinetic energy is at a maximum when the velocity is 16 \displaystyle \text{m}{{\text{s}}^{-1}}.

\displaystyle \text{KE}=\frac{1}{2}\times 0.3\times {{16}^{2}}=38.4\text{ J}

The kinetic energy is at a minimum when the velocity is 6 \displaystyle \text{m}{{\text{s}}^{-1}}.

\displaystyle \text{KE}=\frac{1}{2}\times 0.3\times {{6}^{2}}=5.4\text{ J}

(b) If the ball was projected vertically, it would have a velocity of zero at its maximum height and hence the minimum kinetic energy would be zero. Example 17.2 A ball, of mass 20 grams, hits a wall travelling at 4 \displaystyle \text{m}{{\text{s}}^{-1}} and rebounds at 3 \displaystyle \text{m}{{\text{s}}^{-1}}. Find the energy lost during the bounce.

Solution Find the kinetic energy before the bounce:

\displaystyle \text{KE}=\frac{1}{2}\times 0.02\times {{4}^{2}}=0.16\text{ J}

Find the kinetic energy after the bounce:

\displaystyle \text{KE}=\frac{1}{2}\times 0.02\times {{3}^{2}}=0.09\text{ J}

The loss of energy can then be calculated:

\displaystyle \begin{align} & \text{Energy Lost }=0.16-0.09 \\ & =0.07\text{ J} \end{align}

Example 17.3 A roller coaster has carriages which are modelled as particles of mass 500 kg. A carriage starts at A with speed 12 \displaystyle \text{m}{{\text{s}}^{-1}}. It moves, as shown in the diagram, through the points B, C and D. Find the kinetic energy and speed of the carriage at each of these points.

Solution Find the total energy of the carriage at A.

\displaystyle \begin{align} & \text{Total Energy}=mgh+\frac{1}{2}m{{v}^{2}} \\ & =500\times 9.8\times 1+\frac{1}{2}\times 500\times {{12}^{2}} \\ & =4900+36000 \\ & =40900\text{ J} \end{align}

At the point B:

\displaystyle \begin{align} & 40900=500\times 9.8\times 5+\text{KE} \\ & \text{KE}=\text{4090}0-\text{24500}=16400\text{ J} \\ \end{align}

\displaystyle \begin{align} & 16400=\frac{1}{2}\times 500{{v}^{2}} \\ & v=\sqrt{\frac{16400}{250}}=8.1\text{ m}{{\text{s}}^{\text{-1}}} \\ \end{align}

At the point C there is no potential energy, so the kinetic energy is 40900 J. The speed can then be calculated.

\displaystyle \begin{align} & 40900=\frac{1}{2}\times 500{{v}^{2}} \\ & v=\sqrt{\frac{40900}{250}}=12.8\text{ m}{{\text{s}}^{\text{-1}}} \\ \end{align}

At the point D:

\displaystyle \begin{align} & 40900=500\times 9.8\times 3+\text{KE} \\ & \text{KE}=\text{4090}0-\text{14700}=26200\text{ J} \\ \end{align}

\displaystyle \begin{align} & 26200=\frac{1}{2}\times 500{{v}^{2}} \\ & v=\sqrt{\frac{26200}{250}}=10.2\text{ m}{{\text{s}}^{\text{-1}}} \\ \end{align}

Example 17.4 A bungee jumper has mass 60 kg and uses a rope of length 40m. The jumper comes to rest after falling a distance of 60 m. (a) Find the speed of the jumper when the rope first becomes taut. (b) Find the elastic potential energy stored in the rope, when it is fully stretched.

Solution (a) When the rope becomes taut the bungee jumper has fallen 40 m. The potential energy lost will be equal to the kinetic energy.

\displaystyle \begin{align} & 60\times 9.8\times 40=\frac{1}{2}\times 60{{v}^{2}} \\ & v=\sqrt{\frac{23520}{30}}=28\text{ m}{{\text{s}}^{\text{-1}}} \\ \end{align}

(b) At the lowest point all of the potential energy lost will be equal to the elastic potential energy.

\displaystyle \text{EPE }=60\times 9.8\times 60=35280\text{ J}