Bricklayer problem (involves springs)

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  • #1
lowellite
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Homework Statement


An ingenious bricklayer builds a device for shooting bricks up to the top of the wall where he is working. He places a brick on a verticle compressed spring with force constant k=450N/m and negligible mass. When the spring is released, the brick is propelled upward. If the brick has mass 1.80kg and is to reach a maximum height of 3.6m above the initial position on the compressed spring, what distance must the spring be compressed initially?

Homework Equations


W=Fs
Wtotal=Kf-Ki
F=kx (Hooke's law)

The Attempt at a Solution


Someone else has already solved this problem (but with k=350 instead of 450) https://www.physicsforums.com/showthread.php?t=254633&highlight=bricklayer", though I can't really understand much of it. They said that the total work done on the brick is 0, but I'm not really sure why. Is it because the velocity of the block is 0 when the spring begins to decompress and when the block reaches its maximum height (so the initial and final kinetic energy would both be 0)? For some reason, I can't seem to understand this on a conceptual level.

I'm also not very sure about the rest of the solution. So the work done to compress the spring is W=0.5kd2, where d is how much the spring was compressed. Does that mean when the spring decompresses, an equal amount of work is done?

The work done by gravity from the time when the brick leaves the spring to the time when the brick reaches its maximum height is W=-mgh, where h is the maximum height (3.6m).

Then I'm supposed to relate W=0.5kd2 and W=mgh to get 0.5kd2=mgh, but I don't really understand how this can be done...
 
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  • #2
The total work is 0 because if you consider just potential energy then the potential energy stored in the spring is used to propel the brick into the air; where at the point of K=0 all of the potential energy is then gravitational.

It's easier to think of this purely as conservation of energy. There is an initial point and then a final point. Since we don't consider things like air resistance and energy lost from heat you can just look at time 0 (spring compressed) and time f (3.6m high)
 
  • #3
At the top of its flight all the energy "in" the brick is potential. At the start of its flight, all the energy in the brick is kinetic. Before the start of its flight, all the energy that will go into the brick is potential energy stored in the spring.

RGV
 

1. What is the "Bricklayer problem"?

The "Bricklayer problem" is a physics problem that involves a bricklayer standing on a platform supported by two springs. The springs have different stiffness and are at different distances from the platform. The question is to find the bricklayer's position and the forces on the springs when the platform is in equilibrium.

2. How is this problem related to springs?

The problem is related to springs because the platform is supported by two springs, and the bricklayer's position and forces on the springs are affected by the stiffness and distance of the springs.

3. What factors affect the bricklayer's position and the forces on the springs?

The bricklayer's position and the forces on the springs are affected by the stiffness and distance of the springs, as well as the mass of the bricklayer and the weight of the platform.

4. Is this problem applicable in real-life situations?

Yes, this problem can be applied in real-life situations, such as in engineering and construction, where structures are supported by springs or other types of supports.

5. What is the significance of solving the "Bricklayer problem"?

Solving the "Bricklayer problem" can help in understanding the principles of equilibrium, forces, and energy in a system. It can also be applied in practical situations to optimize the design and stability of structures supported by springs.

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