Finding speed from force-vs-position graph

  • Thread starter Thread starter shaggyace
  • Start date Start date
  • Tags Tags
    Graph Speed
Click For Summary
The discussion revolves around solving a physics problem involving a force-vs-position graph for a 1.7-kg object with an initial speed of 0.44 m/s. Part A was correctly solved, yielding a speed of 0.76 m/s at x=0.99 m. In Part B, confusion arose regarding the object's speed of 0.32 m/s, with attempts to apply kinematic equations leading to incorrect answers. The key insight is that the speed of 0.32 m/s is less than the initial speed, indicating that the object must have been closer to the origin at that speed. Ultimately, the problem was clarified, and the correct interpretation of the motion was achieved.
shaggyace
Messages
11
Reaction score
0

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.
 
Physics news on Phys.org


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 :)
 
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
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
428
  • · Replies 2 ·
Replies
2
Views
4K
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
20
Views
2K
  • · Replies 8 ·
Replies
8
Views
795
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K