String with masses on either end being pulled from the center

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Homework Help Overview

The problem involves two equal masses connected by a string on a frictionless surface, with a force applied at the midpoint. Participants are tasked with determining the change in the center of mass location, the kinetic energy associated with its motion, and the work done by the force over a specified time period.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of center of mass velocity and kinetic energy, with some questioning the treatment of vectors in the context of kinetic energy. There is a focus on the implications of the force's direction and its effect on the overall system's momentum.

Discussion Status

Some participants have provided guidance regarding the need to consider vector components in calculations, particularly in relation to kinetic energy. There is an ongoing exploration of the implications of the applied force and the resulting motion of the system.

Contextual Notes

Participants express uncertainty about the correctness of their calculations and the interpretation of the problem, indicating a need for clarification on the relationship between the components of motion and the overall system behavior.

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Homework Statement


Two equal 1 kg masses are connected by a light 1.6 m string on a flat frictionless surface. A 2 N force is applied to the midpoint of the string which causes the system to accelerate in the positive x-direction.

After 1.5 seconds, the rope makes an angle of 55° with the x-axis.

Determine:
a) The change in the location of the center of mass at time t = 1.5 s
b) The kinetic energy associated with the center of mass motion
c) The work done by the force from time t = 0 to t = 1.5

Homework Equations


W = Fd
Δp = mΔv = Ft
K = 0.5mv2

The Attempt at a Solution


I started with part b and solved for the center of mass velocity.
Ft = mv
v(t) = Ft/m
v(1.5) = (2)(1.5)/2 = 1.5 m/s
K = 2.25 JThen for part a, I integrated my velocity function to get
x(t) = Ft2/2m
x(1.5) = 1.125

For part c, I calculated force displacement to get work
force displacement = 1.125 + 0.8sin(55) = 1.78 m
W = Fd
W = 2 * 1.78
W = 3.56 J

I have no idea if what I did is correct, and I haven't been able to find any similar questions online. Any help is appreciated!
 
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CSPhysics said:
Ft = mv
v(t) = Ft/m
v(1.5) = (2)(1.5)/2 = 1.5 m/s
K = 2.25 J
Force, velocity and momentum are vectors, KE is not. Think about directions. What have you overlooked?
 
haruspex said:
Force, velocity and momentum are vectors, KE is not. Think about directions. What have you overlooked?
I'm thinking of it in terms of a force acting on the entire system and changing its momentum. The force is along the x-axis so the velocity should also be along that axis.

If I needed to calculate the total KE then I would need to consider the y-component of the masses' motion but if I understand it correctly, that should not effect the motion of the center of mass and the KE associated with that motion.
 
CSPhysics said:
I'm thinking of it in terms of a force acting on the entire system and changing its momentum. The force is along the x-axis so the velocity should also be along that axis.

If I needed to calculate the total KE then I would need to consider the y-component of the masses' motion but if I understand it correctly, that should not effect the motion of the center of mass and the KE associated with that motion.
Sorry, you are right. I did not think carefully about what the question was asking.

CSPhysics said:
0.8sin(55)
Cos maybe?
 
haruspex said:
Sorry, you are right. I did not think carefully about what the question was asking.Cos maybe?
No worries!

Yes, that should definitely be cos, Aside from that, does everything else I did see to make sense?
 
CSPhysics said:
No worries!

Yes, that should definitely be cos, Aside from that, does everything else I did see to make sense?
Yes.
 

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