work and kinetic energy

Work and Kinetic Energy




Up to now we have been using kinematics and Newton's laws to describe the motions of objects. These methods work well if the objects are undergoing constant acceleration but they can become extremely difficult with varying accelerations. For such problems, we will find it easier to express the solutions with the concepts of work and kinetic energy.
Goals

• Understand the definition of work.
  
• Determine the kinetic energy of an object.
  
• Understand how work and kinetic energy are related.
  
• Solve problems using the work-energy theorem.
  

Definitions


Kinetic Energy

A scalar quantity which is a function of an object's mass and speed. The kinetic energy of an object is always positive.
Work

To cause a change in the kinetic energy in a moving object due to a net force on that object.
Work

We should start out by saying that in order for there to be a quantity of work there must be a net force and a displacement. The amount of work on done to an object is given by the dot product of the force on the object and its displacement. Mathematically we write work as

If the force is constant then this becomes

The reason we use the dot product is because we are only interested in the component of the force along the displacement. This means that if the force is perpendicular to the displacement then there is no work. The diagram below illustrates this point. Imagine that the two boxes are are on a track so that they may only move from left to right.

The top box will accelerate to the right faster than the bottom one because all of the force is along the direction of motion. The sum of the force applied to the bottom box is unused because it is perpendicular to the direction of motion.
Kinetic Energy

The kinetic energy of an object is defined as

It represents the total amount of work performed to get a moving object from rest to its present velocity.
Work-Energy Theorem

Now that we have a feel for work and kinetic energy, let's see how they are related quantitatively. The equation below shows the work energy theorem. Click here to see how it is derived.

That means that we may describe work as the change in kinetic energy of an object. It also means that if we know an object's kinetic energy at two points, we can find the net work done to it.
Example 1

A block with a mass of 2kg is sliding along a frictionless surface with an initial velocity of 1m/s. A force of 5N is applied parallel to the velocity for 3m. Find the final velocity of the block.

Example 2

A different block with a mass of only 1kg is sliding across the same surface with a velocity of 1.41m/s. If a force of 5N is applied over 3m how does its final kinetic energy compare to the final energy of the block in example 1?

Example 3


A bartender slides a 0.5kg mug of ale down the counter to a customer. He pushes with a force of 3N for 1m then lets it go. If the mug starts with a velocity of 0 and leaves the bartenders hand with a velocity of 2m/s, find the coefficient of friction between the mug and the bar top.




Solutions to Work and Kinetic Energy Problems


Example 1


A block with a mass of 2kg is sliding along a frictionless surface with an initial velocity of 1m/s. A force of 5N is applied parallel to the velocity for 3m. Find the final velocity of the block.

We start by writing down the work-energy theorem:
W = K₂ - K₁


In order to find the final velocity we must first find the final kinetic energy, so we rewrite the above equation as:
K₂ = W + K₁


We get the final kinetic energy by using the definitions for work and kinetic energy.
Now that we know the final kinetic energy, we can solve for the final velocity.
Plugging in the numbers gives
Example 2


A different block with a mass of only 1kg is sliding across the same surface with a velocity of 1.41m/s. If a force of 5N is applied over 3m how does its final kinetic energy compare to the final energy of the block in example 1?

Again, we start by writing down the work-energy theorem:
W = K₂ - K₁


The final energy of the block is:
K₂ = W + K₁


Using the definitions for work and kinetic energy gives us
Which is the same as the block in example 1. Notice that the same change in kinetic energy occurred in both blocks, but the change in velocity was greater with the lighter block.
Example 3


A bartender slides a 0.5kg mug of ale down the counter to a customer. He pushes with a force of 3N for 1m then lets it go. If the mug starts with a velocity of 0 and leaves the bartenders hand with a velocity of 2m/s, find the coefficient of friction between the mug and the bar top.
The first thing we need to do is compute the total work done by the bartender. This comes from the definition of work:
W_b = Fs = (3N)(1m) = 3J


Next, we find how much work he did on the mug. We will use the work-energy theorem for this.
We can then find the work the bartender did against friction by subtracting the work on the mug from the total work.
W_f = W_b - W_m = 3J - 1J = 2J


Reversing the definition for work, we find that the force of friction is
Remembering that the force of kinetic friction is given by we can find that
This is a fairly reasonable number for a coefficient of kinetic friction.