String with masses on either end being pulled from the center

  • Thread starter Thread starter CSPhysics
  • Start date Start date
  • Tags Tags
    Center String
AI Thread Summary
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.
CSPhysics
Messages
3
Reaction score
0

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!
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top