Net Force Calculation Using Velocity Graph

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The discussion focuses on calculating the net force acting on a 5.5 kg object at specific times using its velocity graph. At 3 seconds, the correct force calculation yields 22N, while at 4 seconds, the force is zero. For 7 seconds, the user struggles to determine the acceleration and initially assumes a velocity of 6 m/s, leading to incorrect calculations. Clarification on using the slope of the velocity-time graph reveals that the change in velocity is actually -12 m/s over a 2-second interval, which is crucial for finding the correct acceleration and force. Understanding the slope as rise over run is emphasized as key to solving the problem accurately.
aligass2004
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Homework Statement



http://i241.photobucket.com/albums/ff4/alg5045/Ex5-07.gif

That figure shows the velocity graph of a 5.5 kg object as it moves along the x-axis. What is the net force acting on this object at the given times? a.) 3s b.) 4s and c.) 7s

Homework Equations



F=ma

The Attempt at a Solution



For 3 seconds, I used F= m (change in velocity/change in time), and I got 22N, which was the right answer.

For 4 seconds, I got zero.

I am stuck on how to find the 7 second one.
 
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What is the acceleration at 7s?
 
By looking at the graph, I assumed the velocity was 6m/s. I got the acceleration to be .857 m/s^2. Then I did the same thing that I did for part a, and I got 4.714, but that was wrong.
 
aligass2004 said:
By looking at the graph, I assumed the velocity was 6m/s. I got the acceleration to be .857 m/s^2. Then I did the same thing that I did for part a, and I got 4.714, but that was wrong.

What is the slope of the v-t graph at t = 7s? The slope is the same all the way from 6 to 8s.
 
Yeah, about that slope business. I haven't had a math class using slope for well over three years. I don't remember. I just remember that it's rise over run
 
aligass2004 said:
Yeah, about that slope business. I haven't had a math class using slope for well over three years. I don't remember. I just remember that it's rise over run

Yeah, you can do the first part like that... slope = rise/run = change in velocity/change in time.

Just do it the same way you did the first part... the acceleration is the same all the way from 6s to 8s... what is the change in velocity from 6s to 8s... what is the change in time?
 
The change in velocity is 12 m/s and the change in time is 2 s. So the acceleration is 6m/s^2. Then the force would be 33. I tried that, but it was wrong.
 
aligass2004 said:
The change in velocity is 12 m/s and the change in time is 2 s. So the acceleration is 6m/s^2. Then the force would be 33. I tried that, but it was wrong.

The change is -12m/s.
 
Ohhhhh...thank you for pointing that out.
 

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