An object of mass 19 kg is subjected to 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 .03 seconds. The average of this force is 53.2 Newtons.
Find the change in the velocity of the object in two different ways:
First we find the change in velocity using the acceleration and time interval:
- `dv = 2.8 m/s^2 * .03 seconds = .084 m/s.
Next we find the change in velocity using the impulse and change in momentum:
- `dv = 1.596 kg m/s / ( 19 kg) = .084 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
When the relationship `dv = (Fave `dt) / m is rearranged into the form
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.
- m dv + v dm = F dt.
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