Throwing a frisbee and measuring change in mechanical energy

[sjcorona]

A 75g Frisbee I thrown from a point 1.1 m above the ground with a speed 12 m/s when it has reached a height $.1 m its speed is 10.5 m/s What is the reduction in Emec of the frisbee-earth system because of air drag?


At first I thought Since ΔEmec= Δk+Δu the change is -0.61 J
Then I reasoned that only the kinetic is affected by drag Δk= -1.3 J
Now I'm wondering if I need to split the velocity into x and y components so I can separate the decrease due to gravity and the decrease due to drag? But I don't thinknthats what they want me to do. According to the guide The answer ends up being 0.53 J

 

[Doc Al]

Calculate the total mechanical energy at each point.

 

[sjcorona]

I did that too,Emec2 - Emec1 is still 0.61

 

[Doc Al]

I don't see how you got that answer (or the guide's answer). What is the final height of the frisbee? Is it 0.1 m?

 

[Nickie]

I would approach this problem by first defining the system as the frisbee and the Earth, and identifying the initial and final states of the system. The initial state is when the frisbee is thrown from a height of 1.1 m with a speed of 12 m/s, and the final state is when the frisbee reaches a height of 0.1 m with a speed of 10.5 m/s.

Next, I would consider the different forms of energy present in the system - kinetic energy (KE) and potential energy (PE). The change in mechanical energy (ΔEmec) can be calculated as the difference between the initial and final mechanical energy, which is equal to the change in kinetic energy (Δk) and potential energy (Δu).

In this case, the change in kinetic energy can be calculated as Δk = 1/2 * m * (vf^2 - vi^2), where m is the mass of the frisbee, vf is the final velocity, and vi is the initial velocity. Plugging in the values, we get Δk = -1.3 J.

To calculate the change in potential energy, we can use the equation Δu = m * g * (hf - hi), where m is the mass of the frisbee, g is the acceleration due to gravity, and hf and hi are the final and initial heights, respectively. In this case, the change in potential energy is negligible since the height change is very small (0.1 m). Therefore, we can assume that Δu is approximately equal to 0 J.

As a result, the change in mechanical energy (ΔEmec) is equal to the change in kinetic energy (Δk) since Δu is negligible. Therefore, the reduction in mechanical energy due to air drag is equal to the change in kinetic energy, which is -1.3 J. This means that the frisbee-earth system loses 1.3 J of mechanical energy due to air drag during its flight.

In conclusion, to accurately calculate the reduction in mechanical energy due to air drag, we need to consider both the kinetic and potential energy changes in the system. In this case, the change in mechanical energy is equal to the change in kinetic energy, which is -1.3 J. Therefore, the reduction in mechanical energy due to air drag is 1.3 J.