Set 3 Problem number 10

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object of mass 8.5 kilograms is acted upon by a net force of 46.75 Newtons.

• The object is initially at rest.
• After the first 4.3 seconds, what will the its velocity, and how far will it have traveled?
• What kinetic energy will it attain during the 4.3 seconds?

Solution

The acceleration of an object of mass 8.5 Kg under the influence of a 46.75 Newton force will be a=F/m=( 46.75 Newtons)/( 8.5 Kg)= 5.5 meters per second per second.

• In 4.3 seconds the velocity change will be 4.3( 5.5 m/s) = 23.65 meters per second; since initial velocity is zero this will be the velocity at the end of the 4.3 seconds.
• The average velocity will be the average of this velocity and zero, or ( 23.65 + 0)/2 meters per second = 11.825 meters per second.
• At this average velocity, in 4.3 seconds the object will move 50.84749 meters.
• The kinetic energy will be KE = .5( 8.5 Kg)( 23.65 m/s)^2 = 2377.12 Joules.
• The work done will be the product of the 46.75 Newton force and the 50.84749 meter displacement, or 2377.12 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 kinetic energy attained by the object will be

• 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)