Tag Archives: comparing kinetic energy

College Physics by Openstax Chapter 7 Problem 10

(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.

Solution:

The translational kinetic energy of an object of mass mm moving at speed vv is KE=12mv2KE=\frac{1}{2}mv^{2}.

Part A. The Velocity of the Elephant to have the same Kinetic Energy as the Sprinter

First, we need to solve for the kinetic energy of the sprinter.

KEsprinter=12(65.0 kg)(10.0 m/s)2KEsprinter=3250 J\begin{align*} KE_{\text{sprinter}} & = \frac{1}{2} \left( 65.0\ \text{kg} \right)\left( 10.0\ \text{m}/\text{s} \right)^{2} \\ KE_{\text{sprinter}} & = 3250\ \text{J} \end{align*}

Then, we need to equate this to the kinetic energy of the elephant with the velocity as the unknown.

KEelephant=KEsprinter12(3000 kg)v2=3250 J1500v2=3250v2=32501500v=32501500v=1.4720 m/sv=1.47 m/s  (Answer)\begin{align*} KE_{\text{elephant}} & = KE_{\text{sprinter}} \\ \frac{1}{2}\left( 3000\ \text{kg} \right) v^{2} & = 3250\ \text{J} \\ 1500 v^{2} & = 3250 \\ v^{2} & = \frac{3250}{1500} \\ v & = \sqrt[]{\frac{3250}{1500}} \\ v & = 1.4720\ \text{m}/\text{s} \\ v & = 1.47\ \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. How Larger Energies Needed for the Movement of Larger Animals would Relate to Metabolic Rates

If the elephant and the sprinter accelerate to a final velocity of 10.0 m/s, then
the elephant would have a much larger kinetic energy than the sprinter.
Therefore, the elephant clearly has burned more energy and requires a faster
metabolic output to sustain that speed.   (Answer)\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

College Physics by Openstax Chapter 7 Problem 9

Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h.

Solution:

The translational kinetic energy of an object of mass mm moving at speed vv is KE=12mv2KE=\frac{1}{2}mv^{2}.

The Kinetic Energy of the Truck

For the truck, we are given the following:

m=20000 kgv=110 kmhr×1000 m1 km×1 hr3600 s=30.5556 m/s\begin{align*} m & = 20 000\ \text{kg} \\ v & = 110\ \frac{\text{km}}{\text{hr}} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{hr}}{3600\ \text{s}} = 30.5556\ \text{m}/\text{s} \end{align*}

Substitute these values to compute for the kinetic energy of the truck.

KEt=12mv2KEt=12(20000 kg)(30.5556 m/s)2KEt=9336446.9136 JKEt=9.34×106 J\begin{align*} KE_{t} & = \frac{1}{2} mv^{2} \\ KE_{t} & = \frac{1}{2} \left( 20 000\ \text{kg} \right) \left( 30.5556\ \text{m}/\text{s} \right)^{2} \\ KE_{t} & = 9 336 446.9136\ \text{J} \\ KE_{t} & = 9.34 \times 10^{6} \ \text{J} \end{align*}

The Kinetic Energy of the Astronaut

For the astronaut, we have the following given values

m=80 kgv=27500 kmhr×1000 m1 km×1 hr3600 s=7638.8889 m/s\begin{align*} m & = 80\ \text{kg} \\ v & = 27 500\ \frac{\text{km}}{\text{hr}} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{hr}}{3600\ \text{s}} = 7638.8889\ \text{m}/\text{s} \end{align*}

The kinetic energy of the astronaut is calculated as

KEa=12mv2KEa=12(80 kg)(7638.8889 m/s)2KEa=2334104945.0617 JKEa=2.33×109 J\begin{align*} KE_{a} & = \frac{1}{2} mv^{2} \\ KE_{a} & = \frac{1}{2} \left( 80\ \text{kg} \right) \left( 7638.8889\ \text{m}/\text{s} \right)^{2} \\ KE_{a} & = 2 334 104 945 .0617\ \text{J} \\ KE_{a} & = 2.33 \times 10^{9} \ \text{J} \end{align*}

Comparing the Kinetic Energies of the truck and the astronaut

KEaKEt=2334104945.0617 J9336446.9136 JKEaKEt=250KEa=250 KEt  (Answer)\begin{align*} \frac{KE_{a}}{KE_{t}} & = \frac{2 334 104 945 .0617\ \text{J}}{9 336 446.9136\ \text{J}} \\ \frac{KE_{a}}{KE_{t}} & = 250 \\ KE_{a} & = 250\ KE_{t} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The kinetic energy of the astronaut is 250 times larger than the kinetic energy of the truck.