Set 3 Problem number 12

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object with mass 8.3 kilograms, initially at rest, is acted upon by a force of 53.95 Newtons.

• If the force acts for 4.3 seconds, what will be KE increase of the object, based on its velocity change?
• During the 4.3 seconds, based on the distance it travels, how much work is done by the net force on the object?

Solution

The acceleration of a 8.3 Kg object under the influence of a 53.95 Newton force will be a=F/m=( 53.95 Newtons)/( 8.3 Kg)= 6.5 meters per second per second.

• In 4.3 seconds the velocity will be 4.3( 6.5 m/s^2) = 27.95 meters per second.
• The average velocity will be the average of this velocity and zero, or ( 27.95 + 0)/2 meters per second = `vAve meters per second.
• At this average velocity, in 4.3 seconds the object will move 60.09249 meters.
• The work done will be the product of the distance 60.09249 meters and the 53.95 Newton force, or 3241.99 Joules.
• The kinetic energy will be KE = .5( 8.3 kg)(`speed m/s)^2 = 3241.99 Joules.

Generalized Solution

If an object of mass m, initially at rest, is acted upon by a net force F for time interval `dt, it will experience acceleration a = F / m for time interval `dt. This will result in a velocity change `dv = a `dt = F / m * `dt. Since the object started from rest its final velocity will be

vf = 0 + `dv = F / m * `dt

and its average velocity will be

vAve = (v0 + vf ) / 2 = (0 + F / m * `dt) / 2 = (1/2) (F / m) `dt.

The distance traveled by the object will be

`ds = vAve `dt = (1/2) (F / m) `dt * `dt = (1/2) (F / m) * `dt^2.

The work done on the object will be

`dW = F `ds = F (1/2) (F / m) * `dt^2 = (1/2) F^2 / m * `dt^2. 

Explanation in terms of Figure(s), Extension

The figure below shows the complete relationship between the work done by the net force and the kinetic energy gained by the object.

• If we find `ds as done in previous problems, we can then combine this result with the known force F to obtain the work `dW.
• As determined earlier, if v0 = 0 then `ds = (1/2) (F / m) * `dt^2.
• After a little substitution and simplification we will find that `dW = F `ds = (1/2) F^2 / m * `dt^2.

Figure(s)