Determine Min Work & Final Velocity of Masses in Massless Spring Exp

  • Thread starter Thread starter edud8
  • Start date Start date
AI Thread Summary
The discussion revolves around calculating the minimum work needed to compress a spring between a 1 kg and a 3 kg mass on a frictionless table, with the 3 kg mass moving at 10 m/s after release. The key equation used is the conservation of energy, where the kinetic energy (KE) of the 3 kg mass translates to the work done on the spring. The stored energy in the spring is determined to be 150 J, which corresponds to the total kinetic energy of both masses when released simultaneously. The final velocities of each mass can be calculated using conservation principles, ensuring the results yield neat integers. The focus remains on applying energy conservation to solve for the required parameters without needing to find spring constant or displacement values directly.
edud8
Messages
4
Reaction score
0
1981B2. A massless spring is between a 1 kilogram mass and a 3 kilogram mass as shown above, but is not attached to either mass. Both masses are on a horizontal frictionless table. In an experiment, the 1 kilogram mass is held in place and the spring is compressed by pushing on the 3 kilogram mass. The 3 kilogram mass is then released and moves off with a speed of 10 meters per second.
a. Determine the minimum work needed to compress the spring in this experiment.





The spring is compressed again exactly as above, but this time both masses are released simultaneously.
b. Determine the final velocity of each mass relative to the table after the masses are released.



attempt:
1/2mv^2=1/2kx^2
1/2mv^2=150, but idk how to solve for k or x
 
Physics news on Phys.org
edud8 said:
1/2mv^2=1/2kx^2
1/2mv^2=150, but idk how to solve for k or x
You don't need to. You only need to know the amount of energy stored in the spring, which is the 150 J you found.

This is the total KE of both masses in part 2, and you also need to remember something else that is conserved.

Your answer will be a neat integer.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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