What Did I Do Wrong in My Conservation of Energy Calculation?

AI Thread Summary
The discussion revolves around a conservation of energy calculation involving a 6kg block sliding on a frictionless surface and compressing a spring. The individual initially believed that only kinetic and elastic potential energy were relevant, dismissing gravitational potential energy due to the lack of height. They set up the energy conservation equation but arrived at an incorrect compression value for the spring. Participants pointed out the need to verify the spring constant's units and the accuracy of the potential energy term. The conversation emphasizes the importance of unit consistency in energy calculations.
Maiia
Messages
78
Reaction score
0

Homework Statement


A 6kg block slides on a horizontal frictionless surface with a speed of 1.2 m/s. It is brought momentarily to rest when it hits a bumper that compresses a spring.

Heres a picture of the problem.
springproblem.jpg


When I was doing this problem, I thought there were only two different energies: the Kinetic Energy and the Elastic Potential Energy. I assumed there was no Gravitational PE b/c there was no height involved. Therefore, I assumed (according to Law of Conservation of Energy) that you could set the Kinetic Energy equal to EPE.
(1/2)mv^2 + (1/2)kx^2= (1/2)mv2^2 + (1/2)kx2^2 which reduces to
(1/2)mv^2= (1/2) k(-Y2)
I am using -Y to denote the compression of the spring
if I plug in and solve
(1/2)(6kg)(1.2m/s ^2)= (1/2) (4)(Y2)
I get 1.341640786 m as the amount of compression. However, this answer is wrong. Can someone tell me what I did wrong?
 
Physics news on Phys.org
Maiia said:

Homework Statement


A 6kg block slides on a horizontal frictionless surface with a speed of 1.2 m/s. It is brought momentarily to rest when it hits a bumper that compresses a spring. When I was doing this problem, I thought there were only two different energies: the Kinetic Energy and the Elastic Potential Energy. I assumed there was no Gravitational PE b/c there was no height involved. Therefore, I assumed (according to Law of Conservation of Energy) that you could set the Kinetic Energy equal to EPE.
(1/2)mv^2 + (1/2)kx^2= (1/2)mv2^2 + (1/2)kx2^2 which reduces to
(1/2)mv^2= (1/2) k(-Y2)
I am using -Y to denote the compression of the spring
if I plug in and solve
(1/2)(6kg)(1.2m/s ^2)= (1/2) (4)(Y2)
I get 1.341640786 m as the amount of compression. However, this answer is wrong. Can someone tell me what I did wrong?


Check your units. Spring constant is 4K*N/m
 
Oh wow that was stupid of me. Thanks!
 
Your understanding of the physics is correct, but perhaps you should check if your spring potential energy term with Y_2 is correct. Energy has units of Newton x meter, and the spring constant has units of Newton per meter.
 
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