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Set 4 Problem number 3
A variable force F(t) acts on an object of mass 9 kg for .6 seconds.
- The force vs. clock time graph increases linearly from 0 Newtons at the beginning of the
.6-second interval to 1062 Newtons, then decreases linearly back to 0 at the end of the time
interval..
What is the change in the object's velocity?
To find the change in the object's velocity we first find the impulse of the force.
- The net impulse of a varying force is the area under its force vs. clock time graph.
- Since the force vs. clock time graph in this case will be a triangle whose width is `dt
= .6 seconds and whose altitude is 1062 Newtons.
- The area of this triangle is half the product of its width and altitude, so the impulse
is
- impulse = area under triangle = 1 / 2 (width * altitude) = 1 / 2 ( .6 seconds * 1062
Newtons) = 318.6 Newton seconds.
We now find the change in velocity using the impulse and change in momentum:
- The impulse of the force has been found to be 318.6 Newton Seconds.
- 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 = 318.6 kg m/s / ( 9 kg) = 70.8 m/s.
We first find the impulse of the
force, which is represented by the area under the force vs. clock time graph:
- If the force vs. clock time graph
is triangular, as in this case, the triangle will have a width `dt equal to the duration
of the force and an altitude Fmax corresponding to the maximum force.
- The area under the graph will thus
be
- impulse = 1 / 2 ( `dt * Fmax).
For the present constant-mass
situation we thus have
- `dv = impulse / m = 1 / 2 (Fmax *
`dt) / m.
University Physics Notes:
For very small time subintervals the variable force is nearly constant, and a
Riemann Sum or a trapezoidal approximation will be close to the actual impulse over that
subinterval.
- The sum of the impulses over all subintervals will therefore be very nearly equal
to the actual impulse.
- In the limit as the lengths of the subintervals approach zero, the approximation
becomes precise, and we see that the impulse is indeed given by the integral of the force
function with respect to time.
- The impulse between t = a and t = b is therefore
- impulse = int( F(t), t, a, b)
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Figure description:
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