Conservation of energy stone problem

[colsandurz45]

I have this problem for class and I'm kind of stuck on all of them. Any help would be appreciated.

a stone of weight w is thrown vertically with initial speed v0. air drag force dissipates at a rate of f*y of mechanical energy as the stone travels distance y.
a) show that the maximum height of the stone is

h = (v0)^2/(2g(1+f/w))

b) show that the speed of the stone upon impact is

v = v0*((w-f)/(w+f))^(1/2)

thanks in advance

 

[HallsofIvy]

I have this problem for class and I'm kind of stuck on all of them. Any help would be appreciated.

a stone of weight w is thrown vertically with initial speed v0. air drag force dissipates at a rate of f*y of mechanical energy as the stone travels distance y.

That's poorly written. Force does not "dissapate". I think what is meant is that the air drag force causes the energy to be lost at that rate. Since Energy = force* distance, saying the energy dissapates at rate f*y (y is distance) means that the drag force is the constant f.
The total force on the stone, as it is going up, is F= ma= -mg- f and so the acceleration is a= -g- f/m= -g-fg/w= -g(1+1/w) (w= mg so m= w/g). Integrate that to find the velocity function, integrate the velocity function to find the height function.

a) show that the maximum height of the stone is

h = (v0)^2/(2g(1+f/w))

At its highest point, the velocity of the stone is 0. Solve that equation for t and calculate the height of the rock at that t.

b) show that the speed of the stone upon impact is

v = v0*((w-f)/(w+f))^(1/2)

thanks in advance

Be careful with this one. On the way down, the resistance force is upward: F= ma= -mg+ f so a= -g(1- f/w). Integrate that to find the velocity function (with v(0)= 0 at the top of the throw) and integrate again to find the height function (with h(0)= (v0)^2/(2g(1+f/w))). Set the height= 0 (hitting the ground) and solve for t. Use that t to find the velocity at impact.

Or, perhaps simpler, since the resistance force is constant, find the total energy dissapated and subtract that from the initial kinetic energy. Use that lower kinetic energy to find the velocity on impact.

 

[kyle8921]

I would be happy to help you with this problem. The key concept to understanding this problem is conservation of energy. This principle states that energy cannot be created or destroyed, only transferred or converted from one form to another.

In this problem, the stone is initially given potential energy (due to its height) and kinetic energy (due to its initial speed). As the stone travels, it experiences air drag force, which dissipates some of its mechanical energy. However, the total energy of the stone must remain constant throughout its motion.

To solve part a), we can use the conservation of energy principle and equate the initial energy of the stone to its maximum potential energy at the highest point of its trajectory. This can be written as:

mgh = (1/2)mv0^2 - fy

Where m is the mass of the stone, g is the acceleration due to gravity, h is the maximum height, and fy is the energy dissipated due to air drag.

Rearranging this equation and solving for h, we get:

h = (v0^2)/(2g(1+f/w))

This shows that the maximum height of the stone is dependent on its initial speed, as well as the weight of the stone and the rate at which energy is dissipated due to air drag.

For part b), we can use the same principle and equate the initial energy of the stone to its kinetic energy at the point of impact. This can be written as:

(1/2)mv0^2 = (1/2)mv^2 + fy

Where v is the speed of the stone upon impact. Rearranging and solving for v, we get:

v = v0*((w-f)/(w+f))^(1/2)

This shows that the speed of the stone upon impact is also dependent on its initial speed, as well as the weight of the stone and the rate at which energy is dissipated due to air drag.

I hope this explanation helps you understand the problem better. Remember to always apply the principle of conservation of energy when solving problems involving energy. Good luck with your class!