What is the final velocity of the particle?

AI Thread Summary
The discussion revolves around calculating the final velocity of a 2.40 kg ice block subjected to a force over time. The initial speed is 2.5 m/s, and the net force acting on the block is determined to be 15 N. The correct approach involves using the impulse-momentum theorem, where the area under the force-time graph represents impulse, leading to a change in momentum and thus a change in velocity. The final velocity at t = 11 s is calculated to be 8.8 m/s, contrasting with an erroneous calculation that resulted in 71.25 m/s. The confusion stems from misapplying the equations and not properly accounting for the impulse.
DavidAp
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The image below "gives, as a function of time t, the force component x that acts on a 2.40 kg ice block that can move only along the x axis. At t = 0, the block is moving in the positive direction of the axis, with a speed of 2.5 m/s. What is its velocity, including sign, at t = 11 s?

syqm8z.jpg


Answer: 8.8 m/s"


Relevant equations:
F=ma
Vf = Vo + at

I was thinking that the total area under the graph would give the net force the component seeing that the y-axis represents Force and the x-axis represents Time. With that I got the net force 15N (29N - 14N). So,

F = 25N
F = ma

m = 2.40kg
Vo = 2.5m/s
t = 11
Vf = ?

Therefore, with the given information, I decided to use the following kinematic equation to find acceleration.
Vf = Vo +at
a = (Vf - Vo)/t

So,
F = m(Vf - Vo)/t

And if we solve for Vf we get
Vf = tF/m + Vo

____________________________________________________________
However, when I plug in the units I get an answer that is way way off.
Vf = 11(15)/2.4 + 2.5
Vf = 165/2.4 + 2.5
Vf = 68.75 + 2.5
Vf = 71.25 m/s

I know I added in the units at the end but I checked and the units come out correctly. Therefore, my question is: what did I do wrong? Why am I so far off? Obviously I did something horribly wrong in my calculation, but where?

Thank you for taking the time review my question. Any help is much appreciated.
 
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DavidAp said:
The image below "gives, as a function of time t, the force component x that acts on a 2.40 kg ice block that can move only along the x axis. At t = 0, the block is moving in the positive direction of the axis, with a speed of 2.5 m/s. What is its velocity, including sign, at t = 11 s?

syqm8z.jpg


Answer: 8.8 m/s"


Relevant equations:
F=ma
Vf = Vo + at

I was thinking that the total area under the graph would give the net force the component seeing that the y-axis represents Force and the x-axis represents Time. With that I got the net force 15N (29N - 14N). So,

F = 25N
F = ma

m = 2.40kg
Vo = 2.5m/s
t = 11
Vf = ?

Therefore, with the given information, I decided to use the following kinematic equation to find acceleration.
Vf = Vo +at
a = (Vf - Vo)/t

So,
F = m(Vf - Vo)/t

And if we solve for Vf we get
Vf = tF/m + Vo

____________________________________________________________
However, when I plug in the units I get an answer that is way way off.
Vf = 11(15)/2.4 + 2.5
Vf = 165/2.4 + 2.5
Vf = 68.75 + 2.5
Vf = 71.25 m/s

I know I added in the units at the end but I checked and the units come out correctly. Therefore, my question is: what did I do wrong? Why am I so far off? Obviously I did something horribly wrong in my calculation, but where?

Thank you for taking the time review my question. Any help is much appreciated.

This is a Force time graph. If you factor in the mass, you have an acceleration time graph. The area "under" and acceleration time graph gives change in velocity does it not?
Don't forget it had an initial velocity.

EDIT: Alternately, the area under a Force-time graph gives you the impulse of the force. The impulse gives you the change in momentum, which can easily be changed to a change in velocity.
 
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