Calculating Impulse Using FΔt: A 4.0-N to 15 N Challenge

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To calculate the total impulse imparted to the object, the impulse from each force must be added together. The 4.0-N force acting for 3.0 seconds results in an impulse of 12 Ns (4 N x 3 s), while the 15 N force acting for 1 second contributes an additional 15 Ns (15 N x 1 s). Therefore, the total impulse is 12 Ns + 15 Ns, equaling 27 Ns. The correct answer to the challenge is option d, 27 Ns. Understanding that impulse is additive is crucial for solving similar problems.
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okay I am not sure what the right answer is out of these multiple choices because I am not sure how exactly to use F delta T.

A 4.0-N force acts for 3.0 sec on an object. The force suddenly increases to 15 N and acts for one more second. What impulse was imparted by these forces to the object?

a. 12 Ns
b. 15 Ns
c. 16 Ns
d. 27 Ns

I got like...both A and B as possible answers but I am not sure which one is right..I used F delta T to get 12 because I just did 4 x 3. Then I got 15 by doing 15 x 1. I am not sure how to use the change in time... when i played around with it i was getting numbers that were not any of the choices. Any help is much appreciated.
 
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Impulse is additive. That is, if you have an impulse of A over one period of time, and an impulse of B over another period, then the total impulse over both periods is A+B. (taking into account the direction each time)
 
Thanks! Let's hope I do good on this re-test after I bombed the first momentum test. haha
 
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