The discussion revolves around applying the impulse momentum theorem to determine the final velocity of an object at a specific time, using integral calculus to analyze the forces acting on it.
Some participants have provided guidance on the integration process and the importance of ensuring the correct mode for angle measurements. There is ongoing exploration of the calculations, with differing results being noted, but no consensus has been reached on the final velocity values.
Participants mention specific time constraints and the need for clarity on the calculations leading to the final velocity. There is also a focus on ensuring that the mathematical setup aligns with the physical principles being applied.
You haven't given any details regarding how you came up with -.822 m/s for the final velocity.mpittma1 said:Homework Statement
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Homework Equations
The Attempt at a Solution
I used ∫Fdt = m(vf-vo)
and came up with -.822 m/s for the final velocity...
I have been reworking this problem over and over and cannot come up with a different answer...
Am i wrong in using the impulse momentum theorem?
Your approach is correct. Can you show us how you integrated? What is the anti-derivative of sin(ωt)?mpittma1 said:Homework Statement
https://scontent-a-sjc.xx.fbcdn.net/hphotos-prn2/t1.0-9/10167935_1403417599934459_6123061969742894932_n.jpg
Homework Equations
The Attempt at a Solution
I used ∫Fdt = m(vf-vo)
and came up with -.822 m/s for the final velocity...
I have been reworking this problem over and over and cannot come up with a different answer...
Am i wrong in using the impulse momentum theorem?
Andrew Mason said:Your approach is correct. Can you show us how you integrated? What is the anti-derivative of sin(ωt)?
AM
SammyS said:You haven't given any details regarding how you came up with -.822 m/s for the final velocity.
Why are you trying to find the final velocity anyway ?
Your integral is correct. I can't tell from your answer how you got the 9e-5 value but it is not correct. The given answer is correct. Remember the argument for cos(ωt) is in radians, not degrees.mpittma1 said:sorry, should have done that from the get go, here is what i did: this is for when t = .55 seconds btw
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Andrew Mason said:Your integral is correct. I can't tell from your answer how you got the 9e-5 value but it is not correct. The given answer is correct.
According to your equation:
vf = (1/m)∫Fdt + v0
If you work that out you will get the answer that is given.
AM
mpittma1 said:Worked it out still got v(.55) = -.812 m/s
the answer is suppose to be v(.55) = -.451 m/s
SammyS said:As AM said, ωt is in radians .
Your result looks like your calculator is in degree mode.