HELP: Kinetic Energy Additional Resources

Jet Takeoff

This military jet requires a tremendous amount of work done on it to get its speed up to takeoff speed in the short distance available on the deck of an aircraft carrier.  Some of the work is done by the plane’s own jet engines but work from a catapult is also necessary for takeoff.

Kinetic Energy

Energy is the ability to change an object’s motion or position.  Energy comes in many forms and each of those forms can be converted into any other form.  A moving object has the ability to change another object’s motion or position simply by colliding with it and this form of energy is called kinetic energy.  The kinetic energy of an object can be calculated by the equation

KE = \frac{1}{2} \ mv^2,

where m is the mass of the object and v is its velocity.  The kinetic energy of a moving object is directly proportional to its mass and directly proportional to the square of its velocity.  This means that an object with twice the mass and equal speed will have twice the kinetic energy while an object with equal mass and twice the speed will have quadruple the kinetic energy.

The SI unit for kinetic energy (and all forms of energy) is \text{kg} \cdot \frac{\text{m}^2}{\text{s}^2} which is equivalent to Joules, the same unit we use for work.  The kinetic energy of an object can be changed by doing work on the object.  The work done on an object equals the kinetic energy gain or loss by the object.  This relationship is expressed in the work-energy theorem W_{\text{NET}} = \Delta KE.

Example Problem: A farmer heaves a 7.56 kg bale of hay with a final velocity of 4.75 m/s.

(a) What is the kinetic energy of the bale?

(b) The bale was originally at rest.  How much work was done on the bale to give it this kinetic energy?

Solution:

(a) KE = \frac{1}{2} \ mv^2 = \left(\frac{1}{2}\right)(7.56 \ \text{kg})(4.75)^2 = 85.3 \ \text{Joules}

(b) \text{Work done} = \Delta KE = 85.3 \ \text{Joules}

Example Problem: What is the kinetic energy of a 750. kg car moving at 50.0 km/h?

Solution: \left(\frac{50.0 \ \text{km}}{\text{h}}\right) \left(\frac{1000 \ \text{m}}{\text{km}}\right) \left(\frac{1 \ \text{h}}{3600 \ \text{s}}\right)=13.9 \ \text{m/s}

KE = \frac{1}{2} \ mv^2 = \left(\frac{1}{2}\right)(750. \ \text{kg})(13.9 \ \text{m/s})^2 = 72,300 \ \text{Joules}

Example Problem: How much work must be done on a 750. kg car to slow it from 100. km/h to 50.0 km/h?

Solution: From the previous example problem, we know that the KE of this car when it is moving at 50.0 km/h is 72,300 Joules.  If the same car is going twice as fast, its KE will be four times as great because KE is proportional to the square of the velocity.  Therefore, when this same car is moving at 100. km/h, its KE is 289,200 Joules.  Therefore, the work done to slow the car from 100. km/h to 50.0 km/h is (289,200 \ \text{Joules}) - (72,300 \ \text{Joules}) = 217,000 \ \text{Joules}.

Summary

  • Energy is the ability to change an object’s motion or position.
  • There are many forms of energy.
  • The energy of motion is called kinetic energy.
  • The formula for kinetic energy is KE = \frac{1}{2} \ mv^2.
  • The work done on an object equals the kinetic energy gain or loss by the object, W_{\text{NET}} = \Delta KE.

Practice

The following video discusses kinetic energy. Use this resource to answer the questions that follow.

  1. Potential energy is present in objects that are ______________.
  2. Kinetic energy is present in objects that are ______________.
  3. What formula is given for kinetic energy?

Practice problems involving kinetic energy:

http://www.physicsclassroom.com/Class/energy/u5l1c.cfm

Review

  1. A comet with a mass of 7.85 × 1011 kg is moving with a velocity of 25,000 m/s.  Calculate its kinetic energy.
  2. A rifle can shoot a 4.00 g bullet at a speed of 998 m/s.
    1. Find the kinetic energy of the bullet.
    2. What work is done on the bullet if it starts from rest?
    3. If the work is done over a distance of 0.75 m, what is the average force on the bullet?
    4. If the bullet comes to rest after penetrating 1.50 cm into a piece of metal, what is the magnitude of the force bringing it to rest?

The Physics Classroom - Kinetic Energy

Review how kinetic energy is related to mass and velocity by watching this cartoon:http://www.schooltube.com/video/faddf7cb14ade293baad/

Practice

At the following URL, review kinetic energy and how to calculate it. Then take the quiz at the bottom of the Web page. Be sure to check your answer.http://www.physicsclassroom.com/class/energy/u5l1c.cfm

Key Equations

Kinetic energy

K = \frac{1}{2}mv^2 \begin{cases}m & \text{mass~(in~kilograms,~kg)}\\v & \text{speed~(in~meters~per~second,~}\text{m}/\text{s}\text{)}\end{cases}

Guidance
The energy of motion is kinetic energy, KE. Whenever an object is in motion it has kinetic energy. The faster it is going, the more energy it has.

Example 1

You are using a sling to throw a small stone. If the sling is .5 m long and you are spinning it at 15 rad/s, how high would the rock go if you throw it straight up?

Solution

We'll start by setting the kinetic energy of rock to it's gravitational potential energy at it's maximum height and then solving for the rock's height.

KE_i&=PE_f\\\frac{1}{2}mv^2&=mgh\\h&=\frac{v^2}{2g}\\

We still don't know the rock's linear velocity, but we do know the sling's angular velocity and radius so we can put those into the equation instead.

h&=\frac{(\omega r)^2}{2g}\\h&=\frac{(15\;\text{rad/s} * .5\;\text{m})^2}{2*9.8\;\text{m/s}^2}\\h&=2.9\;\text{m}\\

Time for Practice

  1. A bomb with 8 x 104 J of potential energy explodes. Assume 20% of its potential energy is converted to kinetic energy of the metal pieces flying outward (shrapnel).
    1. What is the total kinetic energy of the shrapnel?
    2. Assume the average mass of the shrapnel is 0.4 kg and that there are 200 pieces. What is the average speed of one piece?
  2. A 1500 kg car starts at rest and speeds up to 3.0 m/s.
    1. What is the gain in kinetic energy?
    2. We define efficiency as the ratio of output energy (in this case kinetic energy) to input energy. If this car’s efficiency is 0.30, how much input energy was provided by the gasoline?
    3. If 0.15 gallons were used up in the process, what is the energy content of the gasoline in Joules per gallon?
  3. An airplane with mass 200,000 \;\mathrm{kg} is traveling with a speed of 268 \;\mathrm{m/s}.
    1. What is the kinetic energy of the plane at this speed?

    A wind picks up, which causes the plane to lose 1.20 \times 10^8 \;\mathrm{J} per second.

    1. How fast is the plane going after 25.0 seconds?

Answers to Selected Problems

  1. a.  1.2 \times 10^6 J  b.  20 m/s
  2. a. 6750 J b. 2.25 \times 10^5 \;\mathrm{J} c. 1.5 \times 10^5 \;\mathrm{J/gallon \ of \ gas}
  3. a. 7.18 \times 10^9 \;\mathrm{J}

After you have completed this part of the lesson, you can check the associated box on the main course page to mark it as complete

Last modified: Tuesday, 19 January 2016, 9:30 AM