Practice Applying Work & Kinetic Energy Formulas

Practice Applying Work & Kinetic Energy Formulas
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  • 0:04 Work and Kinetic Energy
  • 0:40 Important Formulas
  • 1:19 Moving a Dog Across the Floor
  • 4:24 Bench Press Example
  • 5:59 Lesson Summary
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Lesson Transcript
Instructor: Artem Cheprasov
In this lesson, you're going to learn about practical applications of work and kinetic energy formulas. Specifically, we'll look at how these formulas can be applied to moving an object across a surface and lifting weights.

Work and Kinetic Energy

Work and kinetic energy formulas are often utilized by scientists and engineers to solve a variety of problems. You can think of doing work on an object as applying a force along the axis of the object's displacement. For instance, when you go to your favorite store and push a shopping cart across the aisle, you're applying a force in the direction of the cart's motion, thereby doing work on it so to speak.

Furthermore, a moving object contains energy resulting from its motion, known as kinetic energy. In today's lesson, you will learn to apply work and kinetic energy formulas to practical problems.

Important Formulas

The three main formulas you need to keep in mind are as follows:


The top formula states that the work done on an object by an applied force is equal to the applied force times the object's movement distance times the cosine of the angle between the applied force and movement distance.

In the middle formula, we see that kinetic energy is equal to one-half times the object's mass times the square of its velocity. The bottom formula shows that the net work performed is equal to the difference of the final and initial kinetic energies.

Now that you've got that down pat, let's put everything to use with some cool examples.

Moving a Dog Across the Floor

Suppose your dog is lying on the floor and refuses to move. Everyone is allowed to be lazy, dogs included!

Since you really need the dog to be on the other side of the room though, you decide to calculate how much work it would take to move it there in various ways. Assume that it always lies flat on the floor while you are moving it.

Suppose you were to push it from behind. How much work will be done by an applied force of 100 Newtons parallel to the floor, assuming that the floor is frictionless, in moving the dog 17 meters?

Applying the formula work equals force times distance times cosine of theta, we have 100 Newtons times 17 meters times cosine of 0 degrees,


which comes out to be 1700 Joules.

Suppose you tied a rope around its body to pull it across the floor this time around. Assume that the angle between the rope and the horizontal is 25 degrees. How much work is done by the force applied via the rope in pulling the dog 17 meters across a frictionless floor?

Since the rope makes a 25 degree angle with the floor, we now have 100 Newtons times 17 meters times cosine of 25 degrees,


which is 1541 Joules.

Note that if we increase the angle between the applied force and object's direction of motion from 0 to 90 degrees, the work done by this force decreases.

Now, suppose your dog weighs 15 kilograms and the coefficient of friction between the dog's body and the floor is 0.2. How much work is done by the force of friction in pushing the dog 17 meters across the floor?

Recall that the force of friction is calculated by multiplying the friction coefficient, myu, by the normal force, FN . , which in our situation equals myu times m times g. We compute the force of friction to be 29.4 Newtons. Multiplying everything together, or 29.4 Newtons times 17 meters times the cosine of zero degrees


comes out to be 500 Joules.

And here's another twist. Let's say that after you finally brought your 15 kilogram dog to the other side of the room, it got up and started running away from you after you've annoyed the heck out of it by moving it back and forth the entire day. Can't blame the dog, can you?

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