Set 3 Problem number 11

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object, initially at rest, is acted upon by a net force of 7.5 Newtons.

• The object has mass 2.5 kilograms.
• The force acts for 7.3 seconds.
• What velocity will the object attain and how far will it travel during this time?
• How much work is done by the net force on the object?

Solution

The acceleration of a 2.5 Kg object under the influence of a 7.5 Newton net force will be a=F/m=( 7.5 Newtons)/( 2.5 Kg)= 3 meters per second per second.

• In 7.3 seconds the velocity will be 7.3( 3 m/s) = 21.9 meters per second.
• The average velocity will be the average of this velocity and zero, or ( 21.9 + 0)/2 meters per second = 10.95 meters per second.
• At this average velocity, in 7.3 seconds the object will move 79.93501 meters.
• The work done will be the product of the distance 79.93501 m and the 2.5 Newton force, or 599.5125 Joules.
• Note that the KE of the object at 21.9 m/s will be .5 m v^2 = 599.5125 Joules; thus the work done by the net force is equal to KE gained.

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 is thus

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

• The kinetic energy attained by the object will be identical to the work done by the net force:

• KE = .5 m vf^2 = .5 m (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.

• For the purposes of this problem, note that vf is obtained by the same series of relationships as before and ignore everything to the right of v0.
• Note that if we simply combine our knowledge of vf with the originally given mass m, we obtain KE = .5 m vf^2.
• As determined earlier, if v0 = 0 then vf = F / m * `dt, so after a little substitution and simplification we will find that KE = (1/2) F^2 / m * `dt^2.

Figure(s)