Set 3 Problem number 10

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

If an object of mass 4.1 Kg is pushed by a net force of 30.75 Newtons for 6.9 seconds, how far does it travel and what is its final velocity, assuming that no energy is dissipated? How much work is done?

Solution

The acceleration of an object of mass 4.1 Kg under the influence of a 30.75 Newton force will be a=F/m=( 30.75 Newtons)/( 4.1 Kg)= 7.5 meters per second per second.

• In 6.9 seconds the velocity change will be 6.9( 7.5 m/s) = 51.75 meters per second; since initial velocity is zero this will be the velocity at the end of the 6.9 seconds.
• The average velocity will be the average of this velocity and zero, or ( 51.75 + 0)/2 meters per second = 25.875 meters per second.
• At this average velocity, in 6.9 seconds the object will move 178.5375 meters.
• The kinetic energy will be KE = .5( 4.1 Kg)( 51.75 m/s)^2 = 5490.028 Joules.
• The work done will be the product of the 30.75 Newton force and the 178.5375 meter displacement, or 5490.028 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)