Block on Spring without Friction and speed

[Lanc1988]

Homework Statement

A spring is stretched a distance of Dx = 40 cm beyond its relaxed length. Attached to the end of the spring is an block of mass m = 11 kg, which rests on a horizontal frictionless surface. A force of magnitude 35 N is required to hold the block at this position. The force is then removed.

a) When the spring again returns to its unstretched length, what is the speed of the attached object?
b) When the spring has returned only halfway (20 cm), what is the speed of the attached object?

Homework Equations

W = 1/2*m*v^2

The Attempt at a Solution

I have already solved part a. First I calculated the spring constant by 35 = k*0.4 which means k = 87.5

Then I did .5*k*x^2 = .5*m*v^2 and solved for v which is 1.128m/s

For part b I assumed I would just change the x in the above equation to 0.2 instead of 0.4 which would be 0.564 but it is wrong. So I thought the spring constant would change as well so I tried it again changing the k also but it is also not right. What am I doing wrong?

 

[Astronuc]

b) When the spring has returned only halfway (20 cm), what is the speed of the attached object?

Homework Equations

W = 1/2*m*v^2


Then I did .5*k*x^2 = .5*m*v^2

For part b I assumed I would just change the x in the above equation to 0.2 instead of 0.4 which would be 0.564 but it is wrong. So I thought the spring constant would change as well so I tried it again changing the k also but it is also not right. What am I doing wrong?

For part b, the solution tried would be correct if the spring were stretched from its equilibrium halway, but the question asks for the situation where "the spring has returned only halfway (20 cm)", so the spring had been stretch to 0.4 m (from equilibrium or zero force) and returns to 0.2 m (from equilibrium).

 

[Lanc1988]

still not quite sure what I am doing wrong.. the difference between them is still 0.2m.. am I using the wrong equation?

 

[Lanc1988]

would you add the x's then? since its been stretched to 0.4m and then goes back to 0.2m so the x should be 0.6 in the equation i have?

 

[Lanc1988]

anyone? I am sure I've got to be pretty close to the answer but I just must be doing one little thing wrong...

 

[Lanc1988]

can anyone help me with this?.. i don't understand what Astronuc told me to try.. I've been trying to figure out this part for hours now..

 

[Awwnutz]

I don't know if this is a little late, but for your work done by the spring side of the work-energy theorem you need to find how the the block moved between .4m and .2 m. So you'll use (1/2)(87.5)(.4m) - (1/2)(87.5)(.2m) = work done by spring. Use that to set up the rest of the equation and you should be golden. You were calculating how the spring would have moved between .2m and 0m which would cause the block to move at a slower speed.

 

[needhelp4rmu]

I have the same problem...dx=40cm, m=8kg, F=25N
I figured K=62.5
I used what you explained.
(1/2)(62.5)(.4m) - (1/2)(62.5)(.2m)=6.25
don't i plug that in v=sqrt((6.25(.2^2))/8) to find the velocity halfway? i am not getting it right! I probably pluged in something wrong obviously. please help.

 

[Astronuc]

(1/2)(62.5)(.4m) - (1/2)(62.5)(.2m)=6.25

what is this?

The mechanical energy stored in a spring is 1/2 kd², where k is the spring constant and d is the deflection/displacement from equilibrium (i.e. where stored mech energy is zero). Find the energy stored at both deflections d2 and d1, where d2 > d1, then take the difference and equate to the kinetic energy of the mass being accelerated. Remember to use conservation of energy.

Please refer to - http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html#pe2