Set 3 Problem number 11

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object with mass 4.1 kilograms is initially at rest.

• What will be its velocity after being acted upon by a net force of 12.3 Newtons for 8.9 seconds?
• How far will it travel during the 8.9 seconds?
• How much work is done?

Solution

The acceleration of a 4.1 Kg object under the influence of a 12.3 Newton net force will be a=F/m=( 12.3 Newtons)/( 4.1 Kg)= 3 meters per second per second.

• In 8.9 seconds the velocity will be 8.9( 3 m/s) = 26.7 meters per second.
• The average velocity will be the average of this velocity and zero, or ( 26.7 + 0)/2 meters per second = 13.35 meters per second.
• At this average velocity, in 8.9 seconds the object will move 118.815 meters.
• The work done will be the product of the distance 118.815 m and the 4.1 Newton force, or 1461.424 Joules.
• Note that the KE of the object at 26.7 m/s will be .5 m v^2 = 1461.424 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)