Force*Time graph with respect to momentum

by Donnie_b
Tags: forcetime, graph, momentum, respect
Donnie_b is offline
Sep17-07, 02:30 AM
P: 4

If the mass of the object is 3.0 kg, what is its final velocity over the 8.0 s time period?

This is the work I've done so far.

F∆t = m∆v
100 * 8.0 = 3.0 * v
v = 266.67

Now 267m/s seems quite high to me. So I think the problem Im having is that I'm reading the graph incorrectly and extrapolating the incorrect force from it. So basically, can anyone tell me if I have the correct amount of force listed down?
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Kurdt is offline
Sep17-07, 03:46 AM
Sci Advisor
PF Gold
P: 4,975
One approach would be to re-scale the force axis by dividing by the mass of the object and effectively turning it into an acceleration-time graph. The change in speed is then given by the area under the graph.
fizznerd is offline
Sep25-07, 05:04 PM
P: 1
The area under a Force-Time graph is Impulse (equivalent to change in momentum). You can find the area under the curve and that will equal your momentum change. This should allow you to calculate your velocity change.

Couthbaron is offline
Jul9-10, 02:51 PM
P: 2

Force*Time graph with respect to momentum

DonnieB, Fizznerd is right, you need the area under the curve, and you do not need calculus, as that is a trapezoid. THe area of a trapezoid is the average of the bases times the height, which is (4 + 8)seconds/2*100 N = 600 N*s. Set this to mv - mv0, and assuming v0 is zero get v final = 200 m/s.

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