Set 3 Problem number 5

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object originally moving at a constant speed is acted upon for a specified time by a constant force of 260 Newtons.  At the end of the specified time the force is removed and the object proceeds at a new constant velocity.

• If the object traveled a distance of 10 meters while under the influence of the force, and if there was no dissipation of energy, then by how much would the kinetic energy (abbreviated KE) of the object increase?

Solution

The work done on the object will be the product of force and distance, or ( 260 Newtons)( 10 meters) = 2600 Joules.

Since no energy is dissipated, this is equal to the increase in kinetic energy.

Generalized Solution

The idea here is that the work done on the object is equal to its change `dKE in kinetic energy.

• Since the object exerts a force equal and opposite to the force exerted on it, which is the force doing the work to change the object's KE, the object must do work equal and opposite to the work done on it.

When the force exerted on the object is in the direction of the object's motion, it tends to do positive work on the object and thus to speed the object up.

• In this case the force exerted by the object is opposite to the direction of motion and we say that the object does negative work.
• So when the object does negative work, work is actually done on it and it speeds up.

On the other hand if the force exerted on the object is directed opposite to the object's direction of motion it will tend to slow the object down.

• In this case the force exerted by the object is in the direction of motion and the object does positive work, expending kinetic energy in the process (the energy to do the work has to come from somewhere; in this case it comes from kinetic energy).

We say in general that if `dW is the work done by the object (equal and opposite to the work done by the net force acting on the object), then

• `dW + `dKE = 0.

This equation ensures that, for example, when the object does positive work the kinetic energy change must be negative (i.e., the kinetic energy must decrease).

Explanation in terms of Figure(s), Extension

Figure description: The figure below depicts in bar-graph form the relationship between the kinetic energy of an object and the work done by the object.

• The first blue bar depicts the kinetic energy of the object before it does positive work, while the second depicts the kinetic energy after the work is done.
• The red portion of the second bar depicts the work done by the object.
• The total heights of the two bars are identical, showing that the decrease in kinetic energy is equal to the work done by the object.

• Note that when the object does positive work, the force acting on the object does negative work, thereby slowing the object and decreasing its kinetic energy.

This figure depicts the situation when the work done by the object is positive.

• A similar picture for the case where the object does negative work would be harder to construct and to understand, but it is not difficult to understand that when the object does negative work the work done on it is positive, and positive work done on an object increases its kinetic energy.

Figure(s)