Set 4 Problem number 2

An object of mass 8 kg experiences a variable force F(t) (here F(t) indicates function notation, not multiplication of F by t; F(t) is the force at clock time t) for .06 seconds.

If the average force over this time is 83.2 Newtons,

Use Newton's Second Law and your knowledge of uniformly accelerated motion to find the change in the object's velocity.
Use the Impulse-Momentum Theorem to obtain the same result.

Solution

First we find the change in velocity using the acceleration and time interval:

The acceleration of the object will vary with time, but its average will be 83.2 Newtons / 8 kg = 10.4 m/s^2.
During the .06 seconds the velocity of the object will therefore change by

`dv = 10.4 m/s^2 * .06 seconds = .624 m/s.

Next we find the change in velocity using the impulse and change in momentum:

The impulse of the force is the product Fave * `dt = 83.2 Newtons * .06 seconds = 4.992 Newton seconds = 4.992 kg m / s.
This is equal to the change in momentum, by the Impulse-Momentum Theorem.
Since the mass is unchanging the change in momentum is m `dv.
The change in velocity is therefore obtained by dividing the change in momentum by the mass:

`dv = 4.992 kg m/s / ( 8 kg) = .624 m/s.

If a time-dependent force F(t) (note: F(t) does not indicate multiplication of F by t, but rather a functional dependence F(t) on clock time; F(t) is function notation, not multiplication; F(t) is the force at the instant corresponding to clock time t) acts on object of constant mass m for `dt seconds, the object will experience time-dependent acceleration a = F(t) / m for `dt seconds.

For the `dt-second interval, F(t) will have some average value Fave. Thus the force results in velocity change

`dv = a `dt = (Fave / m) `dt = (Fave `dt) / m.

When the relationship `dv = (Fave `dt) / m is rearranged into the form

m `dv = Fave `dt

we have the Impulse-Momentum Theorem for object of constant mass.

We can use the Impulse-Momentum Theorem to find any of the quantities `dv, Fave, m or `dt given the values of three of these quantities.

University Physics Notes:

Using calculus we can prove that d ( mv ) = F dt, which applies to situations in which mass and/or velocity vary.

Since d ( mv ) = m dv + v dm, we have

m dv + v dm = F dt.

When mass is constant, dm = 0 so v dm disappears and we have m dv = F dt, as before.
When velocity is constant we have v dm = F dt, which tells us that to maintain a changing mass at constant velocity requires a force which depends on how quickly the mass is changing (i.e., F = v dm/dt).

Explanation in terms of Figure(s), Extension

Figure description: