How Does a Spring's Compression Relate to Block Height and Work Done?

[InFiNiTeX]

Physic problem... springs

A 700g block is released from rest at height Ho(initial) above a vertical spring with spring constant k=400 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 19.0 cm.

How much work is done
A) by the block on the spring?
B) by the spring on the block?
C) what is the value of Ho?
D) If the block were released from height 2H0 above the spring, what would be the maximum compression of the spring?

If someone could help me figure where to start and the formulas substituted it would be much appreciated!

 

[stunner5000pt]

the work done on the spring is the change i nteh spring's potential energy.
that is given by $$E_{s} = \frac{1}{2} kx^2$$
what is the initial x value and what is the final x value.

the spring did what to the block?? the box fell and it gained _____ and then the spring did something to stop the block. Use energy conservation

$$\Delta K + \Delta U + \Delta E_{s} = 0$$
where K is the kinetic energy and U is the potential energy

you can find the inital velocity Of the block as it hit the spring from that above equation. The work done by the spring is the change in K and U for the block.

now youk now the velocity of the block as it hit the spring.
Now for part 2 again delta K + delta U = 0 this time for the bloc konly give the final velocity, and the initial velocity (read the question), the initial potential energy (unknown) and the final potential energy (since potential is ONLY REFERENCE you can take the final as zero) and you can find Ho

For D just see how Ho affects the spring and work it out. I suggest you use variables for this and not the numbers themselves though it wouldn't make much of a difference

 

[thepatientmental]

To start, we can use the conservation of energy principle, which states that the total energy of a system (in this case, the block and spring) remains constant. In other words, the initial potential energy of the block at height Ho will be equal to the final potential energy of the compressed spring and the block at its maximum compression.

A) The work done by the block on the spring is equal to the change in potential energy of the system. We can use the equation for potential energy, U = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height. In this case, the initial potential energy of the block is mgho and the final potential energy is mghf, where hf is the final height of the block-spring system. Therefore, the work done by the block on the spring is given by W = mgho - mghf.

B) The work done by the spring on the block is equal to the negative of the work done by the block on the spring (as per Newton's third law). Therefore, the work done by the spring on the block is -W = mghf - mgho.

C) To find the value of Ho, we can use the information given in the problem. We know that the block stops after compressing the spring 19.0 cm, which means that the final height of the block-spring system is 19.0 cm. Using the equation for potential energy, we can set the initial potential energy equal to the final potential energy and solve for Ho. So, mgho = mghf, which gives us Ho = hf = 19.0 cm.

D) If the block were released from height 2Ho, the final height of the block-spring system would be 2hf, where hf is the maximum compression of the spring. Using the same equation for potential energy, we can set the initial potential energy (mgho) equal to the final potential energy (mghf) and solve for hf. So, mgho = mghf, which gives us hf = ho/2 = 19.0 cm/2 = 9.5 cm. This means that if the block were released from a height of 2Ho, the spring would compress by 9.5 cm at its maximum.