Maximum Speed and Height of Mass on Spring

[negatifzeo]

Homework Statement

A 3-kg block rests on top of a 2-kg block which is connected to a spring of constant 40 N/m. The upper block is suddenly removed. determine a)The maximum speed of the 2-kg block and b)the maximum height of the 2-kg block.

Homework Equations

This is supposed to be done using work and kinetic energy, and the principle that states Initial KE + Work = Final KE.

The Attempt at a Solution

Initially, the combined mass of the blocks on the spring is 5*9.8=49 N
49=-ky, so the initial y value is -1.23m. I figured the max speed would be at the point where the spring's displacement is 0, either on the way up or on the way down. I also figured the max height would be when the kinetic energy = 0 again. Am I missing anything with the way the problem is setup? The max height value I get isn't very close to the one in the back of the book.

 

[rl.bhat]

Your approach is correct. Show your calculations.

 

[negatifzeo]

Well starting off with max height. We know that the displacement of the spring is initially 1.23m. So assuming that at max height KE=0 and that KE=0 initially,

1/2(40)(1.23^2)-1/2(40)(ymax^2)-(2*9.8)(Ymax+1.23)=0

Solving this for Ymax gets me 1.23m. This obviously isn't true, you would expect the 2kg block to go higher than the original displacement since there was more weight on it initially.

 

[rl.bhat]

1/2(40)(1.23^2)-1/2(40)(ymax^2)-(2*9.8)(Ymax+1.23)=0
This step is wrong.
What is ymax for 2 kg mass? When you remove 3 kg mass, what is the change in PE in the spring?
Hence delta(PE) = KE of 2 kg mass.

 

[negatifzeo]

1/2(40)(1.23^2)-1/2(40)(ymax^2)-(2*9.8)(Ymax+1.23)=0
This step is wrong.
What is ymax for 2 kg mass? When you remove 3 kg mass, what is the change in PE in the spring?
Hence delta(PE) = KE of 2 kg mass.

We haven't covered PE yet in class, but at any rate I don't understand what you are suggesting. The 2*9.8 is from the work done by gravity. Can you clarify?

 

[rl.bhat]

Initially energy stored in the spring is 1/2*k*(ymax)^2
When you remove 3 kg mass, 2 kg mass will move up by a distance y.The total energy is given by
mg(ymax - y) + 1/2 mv^2 + 1/2*k*( y)^2
You can find y by
2*9.8 = -k*y

 

[PhanthomJay]

We haven't covered PE yet in class

I don't know why you haven't studied PE yet, because this problem gets a bit confusing when you try to use the work-energy principle instead of the conservation of energy principle. But nonetheless, if one insists on using this method, be sure to write the correct equation for it, W_total = delta KE, where W_total is the work done by all forces (gravity and the spring, in this case), and delta KE is (KE_final - KE_initial). You've got to be unusually careful with your plus and minus signs when you use this method, and make sure you calculate the work done by the spring correctly.

 

[negatifzeo]

Well, I thought I had done it correctly, PhantomJay. In determining the max height, the initial KE is 0, and the final KE should also be 0, right? It is released from rest once the 3kg mass is lifted. Right? The work in this problem should be given by 1/2kx^2, positive work being done until displacement is zero followed by negative work, plus the negative work done by gravity all throughout. I agree that not using PE is a bit goofy but this is a dynamics class, not physics. (If there is much difference) I'm still confused as to what exactly I am doing wrong.

 

[PhanthomJay]

Well, I thought I had done it correctly, PhantomJay. In determining the max height, the initial KE is 0, and the final KE should also be 0, right? It is released from rest once the 3kg mass is lifted. Right? The work in this problem should be given by 1/2kx^2, positive work being done until displacement is zero followed by negative work, plus the negative work done by gravity all throughout. I agree that not using PE is a bit goofy but this is a dynamics class, not physics. (If there is much difference) I'm still confused as to what exactly I am doing wrong.

After the 2kg mass leaves the spring, there is no more work done by the spring. It leaves the spring when the spring returns to its natural position at 1.23 m from its initially compressed start point. You have an extra term in there.

 

[negatifzeo]

After the 2kg mass leaves the spring, there is no more work done by the spring. It leaves the spring when the spring returns to its natural position at 1.23 m from its initially compressed start point. You have an extra term in there.

Actually for this problem the 2kg block is connected to the spring.

 

[rl.bhat]

When you put 5 kg mass on the spring how much it compresses from its natural position?
How much energy is stored in the spring?
When you leave 2 kg mass on the spring how much it compresses from its natural position?
In this position how much energy is stored in the spring?
What is the change in the energy?
According to the conservation of energy, the decrease in the energy stored in the spring can be seen in the KE of the 2 kg block and rise in the potential energy (mgh) of the block.

 

[PhanthomJay]

Actually for this problem the 2kg block is connected to the spring.

Yeah, sorry, I missed that fact when first reading the problem. In which case, I agree with your equation, but I think your math is off. Not sure about rl.bhat's conservation of energy approach.
EDIT: Per your equation, I get
1/2(40)(1.23^2)-1/2(40)(ymax^2)-(2*9.8)(Ymax+1.23)=0
30.26 -20ymax^2 -19.6ymax-24.1 =0
6.16 - 20ymax^2 -19.6ymax = 0
20ymax2 +19.6ymax - 6.12 = 0
ymax =0.25 m (displacement from initial unstretched position of spring before the masses were added), or
ymax = 1.23 +0.25 = 1.48 m (displacement from initial compressed position of spring before the 3 kg block is instantly removed).

I'm not sure why (or if) you are expecting a higher amount.