A77 g frisbee is thrown from a point 0.99 m above the ground with a speed of 15 m/s. when it has reached a height of 1.3 m, its speed is 12 m/s. what was the reduction in the mechanical energy of the frisbee-earth system because of air drag?

The reduction in the mechanical energy of the Frisbee-Earth = 2.885 J

Explanation:

From the law of conservation of energy,

Mechanical Energy = Ek + Ep

Where Ek = kinetic Energy, Ep = potential Energy.

Change in mechanical energy = change in kinetic Energy + change in potential energy.

ΔM = ΔEk + ΔEp Equation 1.

Reduction in mechanical energy = -ΔM = -(ΔEk + ΔEp)

-ΔM = -[1/2m(v₂²-v₁²) + mg(h₂-h₁)] Equation 2

Where -ΔM = Reduction in mechanical Energy, m = mass of the Frisbee, v₂ = final velocity of the Frisbee, v₁ = Initial velocity of the Frisbee, h₂ = Final height of the Frisbee, h₁ = Initial height of the Frisbee.

Given: m = 77 g = 0.077 kg, v₁ = 15 m/s, v₂ = 12 m/s, h₁ = 0.99 m, h₂ = 1.3 m, g = 9.81 m/s²

Substituting these values into equation 2

-ΔM = -[1/2(0.077)(12²-15²) + 0.077(1.3-0.99)(9.81)]

-ΔM = -[(0.077)(-40.5) + (0.234)]

-ΔM = -[-3.119+0.234]

-ΔM = -[-2.8845]

-ΔM ≈ 2.885 J

Thus the reduction in the mechanical energy of the Frisbee-Earth = 2.885 J

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