Conservation of energy of ball with drag

[simplemail1]

Homework Statement

A .63 kg ball is thrown up with an initial speed of 14 m/s and reaches a maximum height of 8.1m. How much energy is dissipated by the air drag acting on the ball during the ascent?

Homework Equations

KE= 1/2MV^2

The Attempt at a Solution

I'm not very good at physics. I think I have to find the joules by plugging in the numbers accordingly in the equation, but I don't know what to do from there...

KE = .5(.63)(14)^2 = 61.74

The actual answer is -12 J >.<

 

[Dick]

Compare the change in kinetic energy with the change in potential energy. Which is mgh. The difference is the loss to friction.

 

[Moetasim]

As a scientist, it is important to understand the concept of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. In this case, the initial kinetic energy of the ball is converted into potential energy as it reaches its maximum height, and then back to kinetic energy as it falls back down.

However, in the real world, there are always external factors, such as air resistance or drag, that can dissipate some of this energy. In the case of the ball being thrown up, air drag would act in the opposite direction of the ball's motion, slowing it down and dissipating some of its energy.

To calculate the energy dissipated by air drag, we need to use the equation for work, which is W = Fd, where W is work, F is the force applied, and d is the distance over which the force is applied. In this case, the force of air drag is opposite to the direction of motion of the ball, and the distance over which it acts is the height the ball travels.

We can estimate the force of air drag using the equation F = 1/2ρAv^2, where ρ is the density of air, A is the cross-sectional area of the ball, and v is the velocity of the ball. Since we are given the initial velocity and maximum height of the ball, we can use these values to calculate the average velocity of the ball during its ascent.

Using the given values, we get an average velocity of 9.8 m/s for the ball during its ascent. Plugging this into the equation for force, we get F = 1/2(1.2)(0.063)(9.8)^2 = 2.3 N.

Now, we can calculate the work done by air drag using the equation W = Fd. Since the force is constant and opposite to the direction of motion, we can simply multiply the force by the distance traveled, which is the maximum height of the ball, 8.1 m.

Therefore, the work done by air drag is W = 2.3 N * 8.1 m = 18.63 J.

However, since the work done by air drag is in the opposite direction of the ball's motion, it is considered negative. So the energy dissipated by air drag is -18.63 J.

To find the total energy dissipated by air drag during the ball