Max velocity of blocks after spring pushes them apart

AI Thread Summary
The discussion revolves around calculating the maximum velocities of two blocks after being pushed apart by a compressed spring. The spring has a force constant of 3.75 N/m and is compressed by 8.00 cm, with the blocks having masses of 0.250 kg and 0.490 kg. The maximum velocities are affected by varying coefficients of kinetic friction, with specific values provided for analysis. There is confusion regarding the calculated velocities, particularly for the 0.250 kg block, with participants questioning the accuracy of the results. The conversation emphasizes the importance of correctly applying the energy conservation equation to determine the velocities.
joedango
Messages
9
Reaction score
0

Homework Statement


A light spring of force constant 3.75 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.490 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is the following. In each case, assume that the coefficient of static friction is greater than the coefficient of kinetic friction. Let the positive direction point to the right.

(a) For the 0.250 kg block:
At µk = 0.000, the maximum velocity is _____ m/s
At µk = 0.100, the maximum velocity is _____ m/s
At µk = 0.454, the maximum velocity is _____ m/s

(b) For the 0.490 kg block:
At µk = 0.000, the maximum velocity is _____ m/s
At µk = 0.100, the maximum velocity is _____ m/s
At µk = 0.454, the maximum velocity is _____ m/s

Homework Equations



1/2Kx^2=1/2mv^2+fd

The Attempt at a Solution


For µk = 0.000 and m = 0.250 kg, I get:
1/2(3.75)(0.08^2)=1/2(0.25)v^2 + 0
solving for v i get: v=0.309 and v=-0.31
since the 0.250 kg block is moving to the left the answer should be -0.31 m/s, but the answer is incorrect.
 
Physics news on Phys.org
a)
velocity of mass 0.49 is 0.23 m/s
velocity of mass 0.25 is 3.2 m/s
 
how did you obtain those answers? I don't think they're correct.
 
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