Spring and block problem

In summary: The derivation did not give the mass of the second block. You need to use the known mass of the first block to compute the mass of the second block. From F1 = -kx and F2 = -3kx we find that F2 = 3F1 by substitution. F1 is equal in magnitude to the weight of the first block. So, 3F1 = M1g. M1g is the mass of the first block. So, 2M1 = M2.
  • #1
Please help, I can not figure out if I would set up the equation as 3F=-kx

The problem is A 0.70-kg block is hung from and stretches a spring that is attached from the ceiling. A 2nd block is attached to the first one and the amount that the srping stretches from its unstrained length triples. What is the mas of the second block?
 
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  • #2
If F1 is -kx, then F2 is -3kx. F1 is the weight of the first block. F2 is the weight of the two blocks together.
 
  • #3
ok am I using the right equation for this problem, I know I have to find the mass
 
  • #4
Yes, you have the right equation. You just need to relate the forces to one another, as outlined previously, and relate those forces (weights) to the masses of the objects involved.
 
  • #5
F1=F2 and use -kx=-3kx ??
 
  • #6
F1 does not equal F2. From F1 = -kx and F2 = -3kx we find that F2 = 3F1 by substitution
 
  • #8
OK Let's step through it
The first block causes the spring to stretch a distance x. The force the spring provides is proportional to the stretch. It is -kx. The minus sign indicates that the force is in the direction opposite the stretch. Adding the second block triples the distance of stretch, but k stays the same. k is a property of the spring (as long as you don't stretch it too far and ruin it).

There are two cases

F1 = -kx
F2 = -3kx

If you replace the -kx in the second equation with F1 (because they are equal) you get

F2 = 3F1

Now F1 is equal in magnitude to the weight of the first block

F1 = M1g

and F2 is equal to the weight of the two blocks combined

F2 = M1g + M2g

Replce F2 and M1g with their equals

3F1 = F1 + M2g

Subtract F1 from both sides

2F1 = M2g

Replace F1 with its equal M1g

2M1g = M2g

divide both sides by g

2M1 = M2

The mass of the second block is twice the mass of the first block. Use the known mass of the first block to compute the mass of the second block
 
  • #9
i will work through it tonight
 
  • #10
I still can't figure it out
 
  • #11
What do you not understand, the derivation I did to get the result 2M1 = M2, or where to go from there?
 

1. What is the "Spring and block problem"?

The "Spring and block problem" is a classic physics problem that involves a block attached to a spring, with the other end of the spring attached to a fixed surface. The block is either pulled or pushed and then released, causing it to oscillate back and forth due to the spring's restoring force.

2. What are the key variables in the "Spring and block problem"?

The key variables in the "Spring and block problem" are the mass of the block, the spring constant, and the amplitude of the oscillation. These variables determine the frequency, period, and energy of the oscillation.

3. How does Hooke's Law relate to the "Spring and block problem"?

Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. In the "Spring and block problem", this force is responsible for the oscillation of the block and is given by the equation F = -kx, where k is the spring constant and x is the displacement.

4. How does the energy of the "Spring and block problem" change over time?

In the "Spring and block problem", the block's energy changes from kinetic to potential and back again as it oscillates. At the equilibrium position, all of the energy is in the form of potential energy, and at the maximum displacement, all of the energy is in the form of kinetic energy. The total energy remains constant throughout the oscillation.

5. What factors affect the period of the oscillation in the "Spring and block problem"?

The period of the oscillation in the "Spring and block problem" is affected by the mass of the block, the spring constant, and the amplitude of the oscillation. A higher mass or spring constant will result in a longer period, while a larger amplitude will result in a shorter period. The period is also independent of the initial displacement or velocity of the block.

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