Spring potential energy problem

[dorkymichelle]

Homework Statement

A block of mass m = 3.5 kg is dropped from height h = 81 cm onto a spring of spring constant k = 2220 N/m (Fig. 8-37). Find the maximum distance the spring is compressed.

Homework Equations

Pe(spring) = 1/2kx²
Pe(gravity)=mgh
Ke = 1/2mv²

The Attempt at a Solution

so far, i got mgh+1/2mv_i² =1/2mv_f² = 1/2kx²

because the energy used to compress the spring is the amount of energy from the end of the drop.
but I don't have initial velocity or final velocity. or can i use 0 as final velocity at the point that the object hits the spring..

 

[PhanthomJay]

Homework Statement

A block of mass m = 3.5 kg is dropped from height h = 81 cm onto a spring of spring constant k = 2220 N/m (Fig. 8-37). Find the maximum distance the spring is compressed.

Homework Equations

Pe(spring) = 1/2kx²
Pe(gravity)=mgh
Ke = 1/2mv²

The Attempt at a Solution

so far, i got mgh+1/2mv_i² =1/2mv_f² + 1/2kx²

because the energy used to compress the spring is the amount of energy from the end of the drop.
but I don't have initial velocity or final velocity. or can i use 0 as final velocity at the point that the object hits the spring..

The block is dropped from rest, and it ends up momentarily at rest when the spring is fully compressed, not when it first hits the spring. You need to change your PE term to account for the extra height due to the spring displacement in its fully compressed position.

 

[dorkymichelle]

so kinetic initial and kinetic final would both be 0
changing my PE(gravity) term to include the compressed spring,
would the equation I need be mg(h+x) = 1/2kx^2 ? and solve for x to be the length the spring compresses?

 

[PhanthomJay]

Yes, exactly.

 

[rwisz]

I would like to clarify a few things before providing a response. Firstly, it would be helpful to have more context and information about the problem, such as the direction of the block's motion and the orientation of the spring. Additionally, it would be helpful to know if there are any external forces acting on the system, such as air resistance or friction.

Assuming that the block is dropped vertically onto the spring and that there are no external forces, we can use conservation of energy to solve this problem. At the moment the block is dropped, it has potential energy due to gravity, given by mgh. When it reaches the maximum compression of the spring, all of its potential energy will have been converted into elastic potential energy in the spring, given by 1/2kx^2. Therefore, we can set these two energies equal to each other and solve for x, the maximum distance the spring is compressed.

mgh = 1/2kx^2

Solving for x, we get x = √(2mgh/k). Plugging in the given values, we get x = √(2*3.5*9.8*0.81/2220) = 0.18 m. Therefore, the maximum distance the spring is compressed is 0.18 m.

I hope this helps! Just remember to always clarify any missing information and assumptions when solving a problem.