Spring Problem (Need to know if its right)

[Jacob87411]

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 isn't 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

 

[vaishakh]

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

 

[Doc Al]

(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 isn't 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.

 

[vaishakh]

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

 

[Jacob87411]

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?

 

[vaishakh]

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

 

[Jacob87411]

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

 

[Doc Al]

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.

 

[Jacob87411]

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]

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).

 

[Jacob87411]

Ah whoops... .15 not .015...so 1.284 m/s...Thanks