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1 )Here are two questions from the physics online assignment given out by the teacher:
An object of mass 5.5kg (m_{1}) is in equilibrium while connected to a light spring of constant 104 N/m that is fastened to a wall. A second object of mass 5kg is slowly pushed up against mass m_{1}, compressing the spring by the amount 0.29 m.
The system is then released, causing both masses to start moving to the right on the frictionless surface. When m_{1} is at the equilibrium point, m_{2} loses contact with m_{1} and moves to the right with speed v.
Find the value of v.
Answer in units of m/s.
(2) How far apart are the objects when the spring is fully stretched for the first time?
Answer in units of cm.
For question (1), this is what I did:
\SigmaE_{i}=\SigmaE_{f}
1/2*k*l^2=1/2*m*v^2
1/2*(104 N/m)*(0.29 m)^2=1/2*(5.5 kg)(v^2)
v^2=1.59 m^2/s^2
v=1.26 m/s
But when I punched in the answer, it says it is incorrect. I know somehow I did something wrong in the process as my teacher said earlier that this question is tricky.
As I don't really know the velocity m_{2} is traveling upon losing contact with m_{1}, I also have no idea about how to tackle the second question. Some help will really be appreciated!!! ![]()
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for question#1
potential energy of spring when fully compressed = kinetic energy of both masses at equilibrium point
Thus k*x^2/2=(m1+m2)*v^2/2 => v=x*sqrt(k/(m1+m2))
m1=5.5;m2=5;k=104;x=0.29
v=0.91 m/s
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