Force*Time graph with respect to momentum

[Donnie_b]

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 I am 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?

 

[Kurdt]

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]

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]

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.

 

[rwisz]

I would like to clarify a few things about the relationship between force, time, and momentum. Force is the rate of change of momentum, so the area under a force-time graph represents the change in momentum over a certain time period. This means that the units of a force-time graph are not in Newtons (N), but rather in Newton-seconds (N*s).

In your calculation, you have correctly used the equation F∆t = m∆v to find the final velocity of the object. However, the units for force should be in N*s, not just N. So the equation should be (100 N*s) * (8.0 s) = (3.0 kg) * ∆v. This results in a final velocity of 266.67 m/s, which is correct.

It is important to note that the value of force on the graph is not a constant value, but rather the average force over the given time period. So it is possible for the force to vary throughout the 8.0 s time period, but the area under the graph will still represent the total change in momentum.

Additionally, the mass of the object will also affect its final velocity. In this case, the mass of the object is 3.0 kg, but if the mass were to increase, the final velocity would decrease, and vice versa.

In conclusion, your calculation for the final velocity is correct, but it is important to consider the units and the relationship between force, time, and momentum in order to accurately interpret the information from the force-time graph.