Set 3 Problem number 2

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object must gain 14 Joules of energy from a force of 5 Newtons being exerted on it parallel to its direction of motion.

• If the object does not dissipate any of the energy added to it, then over what distance must it be pushed?

Solution

It is assumed that there is no dissipation of energy, so the work done must be equal to the 14 Joules of energy gained from the applied force.

• Since work equals parallel force times displacement, 14 Joules is equal to 5 Newtons multiplied by the distance.
• This could be written as an equation:

• 14 J = ( 5 N)(dist).

• Solving the equation we find that

• dist = 14 J/( 5 N) = 2.8 meters.

Generalized Solution

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

In the present problem, force and displacement are assumed to be parallel and in the same direction.

• In this case, since the work done is equal to the energy expended  we can say that `dW = F `ds.
• Solving we see that F = `ds / `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)