Power as a function of height (gravity and spring involved)

In summary, a "cannon" with a compressed object inside is pointed downwards with a spring. The object has a mass of 50.0kg and the spring has a constant of 49050. The cannon starts at a height of 0 and the object reaches a height of 1 meter. The potential energy at the start is calculated using mgh + 1/2kx^2 and the kinetic energy at the end is 1/2mv^2. To find the amount of power being put out as a function of height, the velocity of the object as a function of height is first solved for. This is given by v(h) = sqrt(-981y^2 - 19.6y +
  • #1
M4573R
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There is a "cannon" pointing downwards with a spring inside where an object is inside and compressed up 1 meter.

mass of object = 50.0kg
k = 49050

height = 0 at bottom of cannon.

Potential energy at start = mgh + 1/2kx^2
Kinetic energy at end = 1/2mv^2

I had to first solve for the object velocity as a function of height 'y' above the bottom of the cannon.

This was: v(h) = sqrt(-981y^2 - 19.6y + 1001)

The question now is to solve for the amount of power being put out as a function of height.

We are given the answer of: p(h) = 49y * v(h), but I am to find this on my own.

I know: P = w*t, w = change in kinetic energy

The change in kinetic energy can be found pretty easily, but what do I do about time?
 
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  • #2
I'm thinking that since the object is starting at a height of 0, and will be ending at a height of 1 meter, then the time it takes to travel that distance is simply the velocity (in m/s) at the end of the cannon.Therefore, I can rewrite P as:P(h) = (1/2mv^2(h) - 1/2mv^2(0)) * v(1)Using the equations for potential energy and kinetic energy given above, we can solve for P(h):P(h) = (1/2 * 50 * [sqrt(-981y^2 - 19.6y + 1001)]^2 - 1/2 * 50 * 0) * sqrt(-981 + 19.6 + 1001)P(h) = 49y * v(h)Therefore, the amount of power being put out as a function of height is: p(h) = 49y * v(h).
 
  • #3


I would approach this problem by first understanding the concept of power and how it relates to the given scenario. Power is the rate at which work is done or energy is transferred. In this case, the power is being generated by the spring inside the cannon as it compresses and releases, causing the object to move.

To find the power as a function of height, we need to consider the forces acting on the object at different heights. At the bottom of the cannon (height = 0), the only force acting on the object is the force of gravity, which is given by Fg = mg. As the object moves up, the spring starts to compress and exert an opposing force on the object. This force, known as the spring force, is given by Fs = kx, where k is the spring constant and x is the displacement of the spring from its equilibrium position.

At any given height, the net force on the object is given by the sum of these two forces, which can be calculated using Newton's Second Law, Fnet = ma. The acceleration of the object can be found using the equation of motion, v(h) = sqrt(-981y^2 - 19.6y + 1001), where y is the height above the bottom of the cannon. Once we have the acceleration, we can find the net force and hence, the power being generated by the spring at that height.

The time component in the equation P = w/t can be found by considering the distance traveled by the object as it moves from the bottom of the cannon to a given height. This distance, which is equal to the height y, can be divided by the velocity of the object at that height to find the time taken.

Therefore, the power as a function of height can be calculated as:

P(h) = Fnet * v(h) = (mg + kx) * v(h) = (mg + kx) * sqrt(-981y^2 - 19.6y + 1001)

Where m is the mass of the object, g is the acceleration due to gravity, and k is the spring constant.

In conclusion, the power as a function of height can be found by considering the forces acting on the object at different heights and the time taken for the object to reach that height. This equation can be used to understand the relationship between power, height, and the properties of the object and the spring involved.
 

1. How does gravity affect power as a function of height?

Gravity plays a significant role in determining the power generated by an object as it moves from one height to another. As an object falls due to gravity, its potential energy decreases and its kinetic energy increases, resulting in a higher power output. This is because the force of gravity accelerates the object, increasing its velocity and kinetic energy.

2. What is the relationship between spring force and power as a function of height?

The power generated by a spring is directly proportional to the amount of force applied to it. As an object attached to a spring moves from one height to another, the spring exerts a force on the object, which in turn generates power. The higher the spring force, the more power is generated.

3. How does the mass of an object affect power as a function of height?

The mass of an object has a direct impact on the power generated as it moves from one height to another. A heavier object will require more force to move, resulting in a higher power output. However, the shape and size of the object can also influence its power output, as these factors affect its air resistance and drag.

4. What factors can affect the efficiency of power as a function of height?

Several factors can influence the efficiency of power generated by an object as it moves from one height to another. These include the object's shape, mass, air resistance, and the amount of force or energy applied to it. Additionally, external factors such as friction and wind can also impact the efficiency of power as a function of height.

5. How can power as a function of height be calculated?

The power generated by an object as it moves from one height to another can be calculated using the formula P = mgh/t, where P is power, m is mass, g is the acceleration due to gravity, h is the height, and t is the time taken to move from one height to another. This formula takes into account the factors of gravity and height, as well as the time it takes for the object to move between the two heights.

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