Register to reply 
Maximum speed of object between 2 springs 
Share this thread: 
#1
Feb2208, 11:56 AM

P: 14

1. The problem statement, all variables and given/known data
A 1.50 kg box moves back and forth on a horizontal frictionless surface between two different springs, as shown in the accompanying figure. The box is initially pressed against the stronger spring, compressing it 4.00 cm, and then is released from rest. (I solved part A, B is: What is the maximum speed the box will reach?) 2. Relevant equations K[tex]_{i}[/tex] + EPE[tex]_{i}[/tex] = K[tex]_{f}[/tex] + EPE[tex]_{f}[/tex] .5mv[tex]^{2}[/tex] [tex]_{i}[/tex] + .5k[tex]_{i}[/tex]x[tex]_{i}[/tex] [tex]^{2}[/tex] = .5mv[tex]^{2}[/tex][tex]_{f}[/tex] + .5k[tex]_{f}[/tex]x[tex]^{2}[/tex][tex]_{f}[/tex] (sorry about the messedup subscripts; I can't make it work right) 3. The attempt at a solution I solved for x[tex]_{}f[/tex] first, obtaining the correct answer of 5.66 cm. I then plugged in values for the other variables: m of 1.5 kg, initial velocity of 0 (because it started from rest?), initial k of 32 N/m, initial x of 4 cm, final k of 16 N/m, final x of 5.66 cm. I converted cm to m and tried to solve for final velocity, but I got .062 m/s and it's supposed to be 1.85 (I got it wrong enough times that it gave me the answer, I just don't know how to solve for it). 


#2
Feb2208, 12:02 PM

HW Helper
P: 2,685

I am not sure exactly how you are trying to solve for the maximum velocity. Can you show your calculations and reasoning? I can't help find your mistake if I can't see it.



#3
Feb2208, 12:05 PM

P: 14

v[tex]_{f2}[/tex] = mv[tex]_{i}[/tex][tex]^{2}[/tex] + k[tex]_{i}[/tex]x[tex]_{i}[/tex][tex]^{2}[/tex]  k[tex]_{f}[/tex]x[tex]_{f}[/tex][tex]^{2}[/tex]
I used the conservation of energy rule to set the initial and final kinetic energy and elastic potential energy equal to each other, then tried to solve for final kinetic energy by subtracting both sides by final elastic potential energy. Then I divided the final kinetic energy by the mass to get final velocity squared. On the other side, that left initial velocity squared (its mass canceled out when I divided by the mass from the other side) plus initial spring constant * inital compression, minus final spring constant * final compression After I plugged in the numerical values stated above, I took the square root to change final velocity squared to just final velocity. 


#4
Feb2208, 12:28 PM

HW Helper
P: 2,685

Maximum speed of object between 2 springs
Your problem is that you have both the potential energy terms involved.
When the block has potential energy in one spring, the other spring is not compressed at all. So, the total energy is either: [tex]TE=1/2k_ix_i^2[/tex] or [tex]TE=1/2k_fx_f^2[/tex] Thus, only one of the above should be involved in the equation you posted above. 


#5
Feb2208, 12:33 PM

P: 14

OH! I feel stupid now... lol. I'll try that and see if I can make my problem work! Thanks.



#6
Feb2208, 12:47 PM

P: 14

The assumption I was relying on was initial velocity is 0... since there were no values aside from initial compression that could be used to get velocity. Is this a correct assumption? Because with a zero velocity, initial kinetic energy is also 0, so that would leave me with final kinetic energy equal to initial elastic potential energy. So initial EPE divided by mass gives final v^2?
edit: oh! I got it right... my last fatal mistake was a unit conversion of cm to m. lol. But I made it! Thank you again for your help. 


Register to reply 
Related Discussions  
Springs: Inclined ramp, object moving down towards spring  Introductory Physics Homework  3  
Object falls, springs catch  Introductory Physics Homework  2  
Maximum Parallel Force of an Object  Introductory Physics Homework  3  
The Speed of a Block (Springs)  Introductory Physics Homework  5  
Does the speed of a wave in a spring depend on the maximum speed of each coil?  Introductory Physics Homework  3 