Question

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

Answer 1

`\Delta E = 8.20 - 4.95 = 3.25 J`

Step-by-step explanation:

Initial total mechanical energy is given as

`ME = U + KE`

here we will have

`U = mgh`

`U = (0.066)(9.81)(1.05) = 0.68 J`

also we have

`KE = (1)/(2)mv^2`

`KE = (1)/(2)(0.066)(15.1)^2`

`KE = 7.52 J`

`ME_i = 0.68 + 7.52 = 8.2 J`

Now similarly final mechanical energy is given as

`U = mgh`

`U = (0.066)(9.81)(1.59) = 1.03 J`

also we have

`KE = (1)/(2)mv^2`

`KE = (1)/(2)(0.066)(10.9)^2`

`KE = 3.92 J`

`ME_f = 1.03 + 3.92 = 4.95 J`

Now change in mechanical energy is given as

`\Delta E = ME_i - ME_f`

`\Delta E = 8.20 - 4.95 = 3.25 J`