Solving Pushing Two Blocks Kinetic Energy

  • Thread starter Thread starter nitrik
  • Start date Start date
  • Tags Tags
    Blocks Two blocks
AI Thread Summary
The discussion centers on a physics problem involving two blocks, A and B, with masses of 4 kg and 1 kg, respectively, being pushed together by a force. The work done on block A is 10 J, and the initial speed of both blocks is 2.0 m/s. The initial kinetic energy calculations yield 18 J for block A and 12 J for block B, but there is confusion among students regarding the final kinetic energy values. It is emphasized that the change in kinetic energy of the entire system must be considered before determining the individual kinetic energies of each block. Understanding the system's total kinetic energy change is crucial for accurate calculations.
nitrik
Messages
2
Reaction score
0

Homework Statement



There are two blocks touching each other, block A, and block B. Block A is being pushed by a force, which in turn moves block A and B at the same time. Block A has a mass of 4kg, block B has a mass of 1kg. (so the force is pushing the larger block, which is pushing the smaller block).

When the hand starts to push, the blocks are moving with a speed of 2.0 m/s. Suppose that the work done by the hand on block A during a given displacement is 10 J. Determine the final kinetic energy of each block



Homework Equations



Change in work (f * d) = change in kinetic energy ( 1/2m(vf)^2 - 1/2m(vi)^2)


The Attempt at a Solution



10J = Kf - 1/2(4kg)(2m/s)^2, Kf = 18J

10J = Kf - 1/2(1kg)(2m/s)^2, Kf = 12J

Now to me, this seems like a pretty straight forward problem. However, my fellow students are getting a different answer. Perhaps I am overlooking something? Any help is appreciated.
Thanks
 
Physics news on Phys.org
Since the blocks move together at the same speeds, you've got to first look at the change in KE of the system (both blocks together) before you calculate the individual final KE of each.
 
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 '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 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