Impulse, I have found a, b and d

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In summary, the magnitude of the net force exerted on a 3.15 kg particle in the x direction varies over time, as shown in the given graph. The impulse of the force is 12.0 N-s, the final velocity of the particle when starting from rest is 3.8 m/s, and the average force exerted on the particle over the time interval from 0 to 5.00 s is 2.4 N. For part (c), the final velocity can be found using the formula v(f) = v(i) + at, where a is the acceleration calculated using F=ma and t is the time interval given.
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snipaj9696
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The magnitude of the net force exerted in the x direction on a 3.15 kg particle varies in time as shown in the figure below.

There is a graph that is supposed to be here. The y-axis is F(N) and the x-axis is t(s)
It starts from the orgin then goes to point (2,4), then to (3,4), then ends at (0,5). I am sorry if this looks confusing, but it wouldn't let me post the graph.

(a) Find the impulse of the force. The answer to this one is 12.0 N-s

(b) Find the final velocity the particle attains if it is originally at rest.
i m/s

(c) Find its final velocity if its original velocity is -3.4 m/s .
i m/s

(d) Find the average force exerted on the particle for the time interval between 0 and 5.00 s.
N

so far I have done..

for part A I found the answer to be 12 (4+4+4)

for part B I did I=mv which came out to v=3.8m/s

I am stuck on part c. I thought I am supposed to use I=m(vf-vi) but I keep getting a wrong answer.

And for part d I used I=f(tf-ti) and found the answer to be 2.4N

So I need part c!
 
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F=ma; a=F/m v(f) = v(i) +at
Do that for each section.
 
  • #3


I would like to commend you for your efforts in solving this problem. Your approach for parts A and B seems correct. For part C, you are correct in using the equation I=m(vf-vi). However, in this case, the initial velocity is not zero, but -3.4 m/s. So the equation becomes I=m(vf-(-3.4)). Solving for vf, we get vf= 6.8 m/s.

For part D, you have correctly used the equation I=f(tf-ti). However, the time interval is not given in the problem. In this case, we can use the area under the curve on the graph to find the impulse, which is also equal to the average force multiplied by the time interval. So, we can write the equation as I= Favg * t or Favg= I/t. Using the given impulse of 12 N-s and the time interval of 5 s, we get an average force of 2.4 N, which matches your answer.

Overall, your approach and calculations seem to be correct. Keep up the good work!
 

1. What is impulse?

Impulse is a term used in physics to describe the change in momentum of an object. It is calculated by multiplying the force applied to an object by the time it is applied.

2. How is impulse related to acceleration?

Impulse and acceleration are related through Newton's Second Law of Motion. The impulse applied to an object is equal to the change in momentum, which is also equal to the mass of the object multiplied by its acceleration.

3. What is the difference between a, b, and d in regards to impulse?

In regards to impulse, a, b, and d are often used to represent different variables. A can represent the force applied to an object, b can represent the time the force is applied, and d can represent the change in momentum of the object.

4. How is impulse measured?

Impulse is measured in Newton-seconds (N*s) or kilogram-meters per second (kg*m/s). It can also be measured in joules (J) as it is the product of a force and a distance.

5. What are some real-life examples of impulse?

Some real-life examples of impulse include a baseball being hit by a bat, a car's airbag deploying in a collision, and a person jumping off a diving board. In each of these scenarios, a force is applied to an object over a certain amount of time, resulting in a change in momentum.

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