Finding speed from force-vs-position graph

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SUMMARY

The discussion centers on calculating the speed of a 1.7-kg object influenced by a force depicted in a force-vs-position graph. The initial speed is 0.44 m/s at an initial position of 0.27 m. The correct speed at position 0.99 m is determined to be 0.76 m/s using the kinematic equation Vf^2 = Vi^2 + 2aΔx, where acceleration was calculated as 0.275 m/s². For the second part, the user initially struggled to find the position where the speed is 0.32 m/s, ultimately realizing that this speed is lower than the initial speed, indicating a need to consider earlier positions, potentially around 0.1 m.

PREREQUISITES
  • Understanding of kinematic equations, specifically Vf^2 = Vi^2 + 2aΔx
  • Knowledge of work-energy principles, particularly W = F(Δx)
  • Basic concepts of force and acceleration in physics
  • Ability to interpret force-vs-position graphs
NEXT STEPS
  • Study the application of kinematic equations in varying scenarios
  • Explore the relationship between work and energy in physics
  • Learn how to analyze force-vs-position graphs for dynamic systems
  • Investigate the implications of negative time in motion problems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding motion dynamics, particularly in relation to force and kinematics.

shaggyace
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Homework Statement



The force shown in the figure (url below) acts on a 1.7-kg object whose initial speed is 0.44 m/s and initial position is x=0.27 m.

http://session.masteringphysics.com/problemAsset/1123629/2/Walker4e.ch07.Pr040.jpg

Part A: Find the speed of the object when it is at the location x=0.99m.

Answer: 0.76 m/s

Part B: At what location would the object's speed be 0.32 m/s?

Homework Equations



Vf^2=Vi^2 + 2aΔx

W=F(Δx)


The Attempt at a Solution



I already answered Part A correctly, using the position/velocity kinematic equation along with the Work=Force x distance equation by using the given mass and the average of the graphed forces through the given distance. For that acceleration, I had 0.275 m/s^2. For Part B, I tried two things, first I used the same kinematic equation with the same acceleration only changed the final velocity to 0.32 m/s. I added the delta x to the 0.27m in the problem and got the wrong answer. Then, I tried setting the initial velocity as that of part A and solving it that way but I still go the wrong answer. I think I am a little confused as to whether the object is slowing down or speeding up and whether or not the problem began after an initial push from rest or is taking place after the object has already been moving. Someone please help me. Thanks.
 
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shaggyace said:

Homework Statement



The force shown in the figure (url below) acts on a 1.7-kg object whose initial speed is 0.44 m/s and initial position is x=0.27 m.

http://session.masteringphysics.com/problemAsset/1123629/2/Walker4e.ch07.Pr040.jpg

Part A: Find the speed of the object when it is at the location x=0.99m.

Answer: 0.76 m/s

Part B: At what location would the object's speed be 0.32 m/s?

Homework Equations



Vf^2=Vi^2 + 2aΔx

W=F(Δx)


The Attempt at a Solution



I already answered Part A correctly, using the position/velocity kinematic equation along with the Work=Force x distance equation by using the given mass and the average of the graphed forces through the given distance. For that acceleration, I had 0.275 m/s^2. For Part B, I tried two things, first I used the same kinematic equation with the same acceleration only changed the final velocity to 0.32 m/s. I added the delta x to the 0.27m in the problem and got the wrong answer. Then, I tried setting the initial velocity as that of part A and solving it that way but I still go the wrong answer. I think I am a little confused as to whether the object is slowing down or speeding up and whether or not the problem began after an initial push from rest or is taking place after the object has already been moving. Someone please help me. Thanks.

Th initial position/velocity/acceleration presumably refers to what was happening when we chose to look at the object. [After all, if we wanted to analyse the motion of cars on a freeway, it doesn't mean none of them are allowed to move until we look at them].

The speed of 0.32, being smaller that the "initial" speed is 0.44 presumably means the answer sought is a negative time, when the object was closer to the origin ; perhaps at 0.1m or a tiny fraction less.
 
Figured it out. Thanks for your help :)
 

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