# Spring Problem (Need to know if its right)

A glider of mass 0.150 kg moves on a horizontal frictionless air track. It is permanently attached to one end of a massless horizontal spring, which has a force constant of 9.0 N/m both for extension and for compression. The other end of the spring is fixed. The glider is moved to compress the spring by 0.180 m and then released from rest.

(a) Calculate the speed of the glider at the point where it has moved 0.180 m from its starting point, so that the spring is momentarily exerting no force.

So potential = kinetic energy
1/2kx^2=1/2mv^2
v^2=kx^2/m
v^2=(9)(.18^2)/.15
v=1.39

(b) Calculate the speed of the glider at the point where it has moved 0.250 m from its starting point.

Here x=.07 because isnt x the distance from the equilibrium point, which is at .18? If so, use the same equation

1/2kx^2=1/2mv^2
v^2=(9)(.07^2)/.15
v=.542 m/s

Thanks for checking

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a - it is given the question itself that it is mommentarily at rest.

Doc Al
Mentor
Jacob87411 said:
(a) Calculate the speed of the glider at the point where it has moved 0.180 m from its starting point, so that the spring is momentarily exerting no force.

So potential = kinetic energy
1/2kx^2=1/2mv^2
v^2=kx^2/m
v^2=(9)(.18^2)/.15
v=1.39
This is correct (but don't leave off the units). Note that the total energy at the point in question is purely KE; the elastic PE is zero at that point. Initial total energy = final total energy.

(b) Calculate the speed of the glider at the point where it has moved 0.250 m from its starting point.

Here x=.07 because isnt x the distance from the equilibrium point, which is at .18? If so, use the same equation

1/2kx^2=1/2mv^2
v^2=(9)(.07^2)/.15
v=.542 m/s
Careful! "PE = KE" is not true in general. What is true is "initial total energy = final total energy". Set the total initial energy (which is the same as in part (a)) equal to the sum of PE + KE at the point in question.

Sorry - I am really frustrated with the misreadings I make.

So in part b:
Initial = Final energy
Initial = 1/2mv^2 = .5 * .15 * 1.39^2 = .1449 J
Final = KE + PE
KE=1/2 * .015 * v^2
PE=1/2kx^2=.5*9*.07^2

Is that the right set up?

Initially KE = 0 and there is only PE. The object is at rest initially.

Oh so PE=1/2kx^2=.5*9*.18^2?

Doc Al
Mentor
Jacob87411 said:
Oh so PE=1/2kx^2=.5*9*.18^2?
Right. That's the initial PE (from the starting point point where it was released); since it was released from rest, that's also the total energy.

Alright, so I did it out..

Initial Energy=Final Energy
Initial energy is stated above
Final Energy=KE+PE=
.5*9*.18^2 = (.5*.015*V^2)+(.5*9*.07^2)

Solving for V gives V=4.06 m/s?

Doc Al
Mentor
Jacob87411 said:
Alright, so I did it out..

Initial Energy=Final Energy
Initial energy is stated above
Final Energy=KE+PE=
.5*9*.18^2 = (.5*.015*V^2)+(.5*9*.07^2)

Solving for V gives V=4.06 m/s?
Better check your arithmetic since the velocity you solved for in part (a) was the maximum possible (all of the energy was KE in that case). Ah whoops... .15 not .015...so 1.284 m/s...Thanks