What is the final velocity of the particle?

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SUMMARY

The final velocity of a 2.40 kg ice block, initially moving at 2.5 m/s, is calculated using the force-time graph provided. The net force acting on the block is determined to be 15 N, leading to an acceleration that can be derived from the equation F = ma. The correct application of the kinematic equation Vf = Vo + at reveals that the final velocity at t = 11 s is 8.8 m/s, not the erroneous 71.25 m/s calculated by the user. The mistake lies in misapplying the impulse-momentum theorem and not correctly accounting for the initial conditions.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with kinematic equations (Vf = Vo + at)
  • Knowledge of impulse and momentum concepts
  • Ability to interpret force-time graphs
NEXT STEPS
  • Study the relationship between impulse and momentum in physics
  • Learn how to analyze force-time graphs for calculating net forces
  • Explore kinematic equations in-depth, focusing on their applications in various scenarios
  • Practice problems involving mass, force, and acceleration to solidify understanding
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Students studying physics, educators teaching mechanics, and anyone interested in understanding motion and forces in a practical context.

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.
 
Last edited:

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