A 75 g Frisbee is thrown from a point 1.0 m above the ground with a speed of 14 m/s. When it has reached a height of 2.0 m, its speed is 10.1 m/s. What was the reduction in Emec of the Frisbee-Earth system because of air drag?

Answer :

Explanation:

It is given that,

Mass of Frisbee, m = 75 g = 0.075 kg

It is thrown 1 meter above the ground, [tex]h_1=1\ m[/tex]

Speed with which it is thrown, [tex]v_1=14\ m/s[/tex]

When it has reached a height of 2.0 m, [tex]h_2=2\ m[/tex], its speed is 10.1 m/s, [tex]v_2=10.1\ m/s[/tex]

We know that the mechanical energy is equal to the sum of kinetic energy and the potential energy.

The change in kinetic energy,

[tex]\Delta K=\dfrac{1}{2}m(v_2^2-v_1^2)[/tex]

[tex]\Delta K=\dfrac{1}{2}\times 0.075(10.1^2-14^2)[/tex]

[tex]\Delta K=-3.524\ J[/tex]

The change in potential energy,

[tex]\Delta P=mg(h_2-h_1)[/tex]

[tex]\Delta P=0.075\times 9.8\times (2-1)[/tex]

[tex]\Delta P=0.735\ J[/tex]

So, the reduction in mechanical energy is given by :

[tex]E_{mec}=-3.524+0.735=-2.789\ J[/tex]

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