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A 66 g66 g ball is thrown from a point 1.05 m1.05 m above the ground with a speed of 15.1 ms/15.1 ms . When it has reached a height of 1.59 m1.59 m , its speed is 10.9 ms/10.9 ms . What was the change in the mechanical energy of the ball-Earth system because of air drag

Respuesta :

Answer:

[tex]\Delta E = 8.20 - 4.95 = 3.25 J[/tex]

Explanation:

Initial total mechanical energy is given as

[tex]ME = U + KE[/tex]

here we will have

[tex]U = mgh[/tex]

[tex]U = (0.066)(9.81)(1.05) = 0.68 J[/tex]

also we have

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2}(0.066)(15.1)^2[/tex]

[tex]KE = 7.52 J[/tex]

[tex]ME_i = 0.68 + 7.52 = 8.2 J[/tex]

Now similarly final mechanical energy is given as

[tex]U = mgh[/tex]

[tex]U = (0.066)(9.81)(1.59) = 1.03 J[/tex]

also we have

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2}(0.066)(10.9)^2[/tex]

[tex]KE = 3.92 J[/tex]

[tex]ME_f = 1.03 + 3.92 = 4.95 J[/tex]

Now change in mechanical energy is given as

[tex]\Delta E = ME_i - ME_f[/tex]

[tex]\Delta E = 8.20 - 4.95 = 3.25 J[/tex]

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