Energy of block hitting a spring

[Kot]

Homework Statement

A block released from height h hits a spring with constant k, the box compresses the spring a distance x then returns to the same height.

Find the maximum speed of the block.

Homework Equations

K=1/2mv²
U=mgh

The Attempt at a Solution

This solution was provided by my professor but I have a couple of questions about it.

The equilibrium point is S_equ =mg/k
so this is where potential energy is it's lowest and kinetic energy is at it's highest.

1/2mv_max²=mg(h+S_equ) - 1/2k(S_equ)²

mg(h+mg/k) - 1/2k (mg/k)²

1/2 mv_max² = mgh + 1/2 (mg/k)²/k

v_max² = 2gh + (m/k)g²

v_max = √(2gh + (m/k)g²)


I am confused as to why my professor set S_equ equal to mg/k. Did this come from F_spring = kx? Could someone explain this to me?

 

[Tanya Sharma]

Hi Kot...

The block accelerates in the downward direction till it reaches the equilibrium position.After that it decelerates ,till its speed becomes zero(the spring achieves maximum compression) .At the equilibrium position the net force on the block in vertical direction is zero .

Now,the forces are mg downwards and kx(spring force) upwards.So mg-kx=0 or x=mg/k where x is the equilibrium position.

 

[Kot]

I see, so for 1/2mv_max²=mg(h+S_equ) - 1/2k(S_equ)²

The left hand side is the kinetic energy of the block, the right hand side is the potential energy of the block and the spring? Since the spring is compressed there is potential energy stored in it but why is it subtracted from the potential energy of the block?

 

[Tanya Sharma]

I see, so for 1/2mv_max²=mg(h+S_equ) - 1/2k(S_equ)²

The left hand side is the kinetic energy of the block, the right hand side is the potential energy of the block and the spring? Since the spring is compressed there is potential energy stored in it but why is it subtracted from the potential energy of the block?

Using energy conservation,

Loss in potential energy of the block = Gain in kinetic energy of the block + Gain in potential energy of the spring.

mg(h+S_equ) = 1/2mv_max² + 1/2k(S_equ)²

Rearranging,1/2mv_max²=mg(h+S_equ) - 1/2k(S_equ)²

which is what your professor has done.

 

[Kot]

If there were no spring, then the conservation of energy would mean that: loss in potential energy of the block = Gain in kinetic energy of the block, correct?

 

[Tanya Sharma]

Correct

 

[Kot]

Wait a minute, how can I tell that mg(h+S_equ) is loss in potential energy instead of gain and the other terms are gain in kinetic and potential energy? Is there a way to determine this?

 

[Tanya Sharma]

Wait a minute, how can I tell that mg(h+S_equ) is loss in potential energy instead of gain and the other terms are gain in kinetic and potential energy? Is there a way to determine this?

When an object is lifted up from ground to an elevated position ,does the potential energy increase or decrease ?

 

[Kot]

Lifting an object increases it's potential energy since the height increased. So since the distance of the block is h+S_equ the distance is actually lower thus resulting in loss of potential energy?

 

[Tanya Sharma]

PE = mgh where h is the height from the reference point .As the object moves up ,PE increases.As it goes down PE decreases.

 

[Kot]

If h is the distance from a reference point and we add S_equ then that means the height would have increased, and potential energy would have increased... I'm confused :(

 

[Tanya Sharma]

Reference point means h=0(or PE=0).

If you consider equilibrium position to be the reference point i.e measure distances from the equilibrium point ,then the block is initially at height h+mg/k or has PE mg(h+mg/k) .

Now when the block falls down and reaches the equilibrium position(PE=0) ,all this stored PE converts into the Kinetic energy of the block + elastic potential energy of the spring .

Does this help?

 

[Kot]

Yes! Thank you for explaining this to me :)