# Block dropped onto vertical spring, finding work/speed/height

• cupcaked
In summary, the conversation is about a 0.25kg block being dropped onto a vertical spring with a spring constant of 250 N/m. The block becomes attached to the spring and compresses 0.12m before coming to rest. The work done by the force of gravity while the spring is being compressed is 0.294 J, and the work done by the spring is 1.8 J. To find the speed of the block just before it hits the spring, the conservation of mechanical energy equation is used, resulting in v = \sqrt{19.6h}. To determine the height from which the block was released, the potential energy of the block is equated with the energy stored in the spring and solved
cupcaked
I need some help with this problem, to check if my answers and formulas are right/how to solve the last parts...

A 0.25kg block is dropped onto a vertical spring with spring constant k= 250 N/m. The block becomes attached to the spring and the spring compresses 0.12m before coming momentarily to rest.

a) While the spring is being compressed what is the work done by the force of gravity?
W = Fdcosθ
W = (9.8 * 0.25)(0.12)(cos0)
W = 0.294 J

b) What is the work done by the spring?
W = -1/2kx^2
W = -1/2(250)(0.12)^2
W = 1.8 J

c) What was the speed of the block just before it hits the spring?
ME = KE + PE
MEi = (1/2)(0.25)(0 + (0.25)(9.8)(h)
MEi = 2.45h

MEf = (1/2)(0.25)(v^2) + (0.25)(9.8)(0)
MEf = 0.125v^2

MEi = MEf
2.45h = 0.125v^2
v^2 = 19.6h
v = $\sqrt{19.6h}$ ...??

d) From what height was the block released relative to the top of the spring?
I'm not sure how to do this, maybe I did the last part wrong or I keep overlooking something stupid...?

Fall in PE of the block = mg(h+x)
Energy stored in the spring = 1/2*k*x^2
Equate them and solve for h.

## 1. How do you calculate the work done when a block is dropped onto a vertical spring?

The work done in this scenario can be calculated by multiplying the force of gravity acting on the block by the distance it falls, which is equal to the height of the spring. This can be represented by the equation W = mgh, where W is the work done, m is the mass of the block, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the spring.

## 2. How can you find the speed of the block after it is dropped onto the vertical spring?

The speed of the block can be found by using the conservation of energy principle. The potential energy of the block at the top of the spring is equal to the kinetic energy of the block at the bottom of the spring. This can be represented by the equation mgh = 1/2 mv², where m is the mass of the block, g is the acceleration due to gravity, h is the height of the spring, and v is the speed of the block.

## 3. What is the relationship between the height of the spring and the work done on the block?

The height of the spring is directly proportional to the work done on the block. This means that as the height of the spring increases, the work done on the block also increases. This can be seen in the equation W = mgh, where h is the height of the spring. Therefore, the higher the spring, the more work is done on the block when it is dropped onto the spring.

## 4. Can you determine the height of the spring if the mass of the block and the work done on it are known?

Yes, the height of the spring can be determined using the equation h = W/mg, where W is the work done on the block, m is the mass of the block, and g is the acceleration due to gravity. This equation can be rearranged to h = v²/2g, where v is the speed of the block. Therefore, if the work done and/or the speed of the block are known, the height of the spring can be calculated.

## 5. What factors can affect the speed of the block when it is dropped onto the vertical spring?

The speed of the block can be affected by various factors such as the mass of the block, the height of the spring, and the presence of air resistance. Additionally, the material and stiffness of the spring can also have an impact on the speed of the block. Other external factors such as friction and external forces can also affect the speed of the block.

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