Set 3 Problem number 1

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object is moving too fast, and is trying to slow down.

• It is pushing against another object in front of it, which is acting to slow it down.
• It pushes with a force of 5 Newtons.
• How much work does it do on the other object over a distance of 14 meters?
• How much energy of motion does it therefore lose to the other object?

Solution

Work is the product of parallel force and displacement. In this case the force is directed against an object in front of the first object, and must therefore be in the direction of motion.  The force is therefore parallel to the displacement.

• The force is 5 Newtons and displacement is 14 meters.
• The work is therefore the product ( 5 Newtons)( 14 meters) = 70 Joules.

Generalized Solution

The precise definition of the work done by a force is the product of the force component  parallel to motion and the displacement through which the force acts.

• In the present problem,  force and displacement are assumed to be parallel.
• In this case we can say that `dW = F `ds.
• We  say that in order to perform this work `dW, we must expend energy `dW:

• The work done and  the energy required to do the work are equivalent.

• Usually in the process of doing work `dW, the process is not 100% efficient and energy in excess of `dW is required.  In no case can the energy required by less than `dW.

Explanation in terms of Figure(s), Extension

The figure below depicts a force F acting on a moving object along a straight-line path from  point A to point B.

• The force is considered uniform in the direction from A to B.
• The work  done by the force is `dW = F `ds. 

Figure(s)