Set 3 Problem number 8

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object with mass 8.5 kilograms is initially at rest.

• What will be its acceleration when acted upon by a net force of 63.75 Newtons?
• What speed will it attain as a result of this force in 2.9 seconds?
• How far will it travel during the 2.9 seconds?

Solution

The acceleration of an object of mass 8.5 Kg under the influence of a 63.75 Newton force will be

• a=F/m=( 63.75 Newtons)/( 8.5 Kg)= 7.5 meters per second per second.

In 2.9 seconds the velocity change will be

• `dv = 2.9( 7.5 m/s) = 21.75 meters per second.

Since the object starts from rest this will also be the final velocity.

The average velocity will be the average of this final velocity and zero, or ( 21.75 + 0)/2 meters per second = 10.875 meters per second.

At this average velocity, in 2.9 seconds the object will move 31.5375 meters.

Generalized Solution

The rate at which velocity changes, or the acceleration, is a = F / m.

• During a time interval `dt, the velocity will therefore change by a `dv = (F / m) `dt.
• If the object starts from rest its final velocity will therefore be 0 + `dv = (F / m) `dt.
• The average velocity will be the average of its initial and final velocities, or (F / (2m) ) * `dt.
• The distance moved will therefore be

• `ds = vAve * `dt = (F / (2m) ) * `dt².

Explanation in terms of Figure(s), Extension

The first figure below depicts a 'flow' triangle for an object of mass m subjected to a force F.

• The relationship a = F / m can be understood as saying that greater force implies proportionally greater acceleration for a given mass while greater mass implies proportionally less acceleration for a given force.
• The relationship F = m * a can be understood as saying that to achieve a given acceleration a greater force must be exerted on a greater mass and that for a given mass a greater acceleration will require a greater force.
• The relationship m = F / a tells us that for a given observed acceleration the greater the force the greater the mass being accelerated, and for a given applied force a greater acceleration implies that less mass is being accelerated.

The second figure depicts a 'flow' diagram for an object of mass m, initially with velocity v0, subjected to a force F for time interval `dt.

• From the force and the mass we deduce the acceleration (the 'blue' triangle).
• From the acceleration and the time interval we deduce the change `dv in velocity (the 'green' triangle).
• From the change in velocity and initial velocity we obtain the final velocity (the 'purple' triangle).
• From initial and final velocities we obtain average velocity (the 'red' triangle) and from average velocity and `dt we obtain the displacement `ds.

Figure(s)

Figure(s)