Impulse Chapter 9: Net Force & Velocity Solutions

  • Thread starter Thread starter snipaj9696
  • Start date Start date
  • Tags Tags
    Impulse
Click For Summary
The discussion focuses on solving problems related to impulse and velocity for a particle subjected to a varying net force. The impulse calculated from the force graph is 12.0 N-s. For part (b), the final velocity of the particle from rest is determined to be 3.8 m/s. The participant struggles with part (c), attempting to apply the impulse-momentum theorem but encounters errors in their calculations. For part (d), the average force over the time interval is found to be 2.4 N, but clarification is needed for part (c).
snipaj9696
Messages
12
Reaction score
0
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
 
Physics news on Phys.org
In order to get help at this forum, you have to show some work.

What have you done on this problem? What have you tried so far? If you are stumped at the beginning, then tell me what your thoughts on the problem are. What concepts/ formulas apply?
 
Any attempts/ideas?
 
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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
7
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
Replies
6
Views
2K
Replies
19
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
10
Views
1K
  • · Replies 28 ·
Replies
28
Views
9K
Replies
2
Views
2K