String with masses on either end being pulled from the center

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A system of two equal 1 kg masses connected by a 1.6 m string is accelerated by a 2 N force applied at the midpoint, resulting in a 55° angle with the x-axis after 1.5 seconds. The center of mass velocity was calculated as 1.5 m/s, leading to a kinetic energy of 2.25 J. The change in the center of mass location was determined to be 1.125 m. The work done by the force during this time was calculated to be 3.56 J, although there was confusion regarding the direction of components in the calculations. Overall, the approach to solving the problem was mostly correct, with adjustments needed for vector components.
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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.
 
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?

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