You drop a box. mass m = 3.0 kg. from 10 meters above the ground. The initial sp

Question

You drop a box. mass m = 3.0 kg. from 10 meters above the ground. The initial speed is zero.Now because of air drag, the box reaches a terminal speed of 6 1 m/s, when ireaches 5 meters above ,the ground So.....the speed is 6.1 m/s from 5 meters down to zero meters,just before it hits the ground.Obtain the work done by gravity, from 10 meters to 5 meters.Obtain the change m KE from 10 meters to 5 meters. Using the work-kinetic energy theorem, obtain the work done by air drag from 10 meters to 5 meters. Obtain the work done by gravity, from 5 meters to 0 meters. Obtain the change in KE from 5 meters to 0 meters.

Explanation / Answer

Here ,

mass , m = 3 Kg

height , h1 = 10 m

h2 = 5m

speed , u = 6.1 m/s

a) work done by gravity = m * g *(h2 - h1)

work done by gravity = 3 * 9.8 * (10 - 5)

work done by gravity = 147 J

b)

change in kinetic energy= final kinetic energy - initial kinetic energy

change in kinetic energy = 0.5 * m * u^2 - 0

change in kinetic energy = 0.5 * 3 * 6.1^2 J

change in kinetic energy = 55.82 J

c)let the work done by air drag is W

work done by gravity + Wd = change in kinetic energy

147 + Wd = 55.82

Wd = -91.2 J

the work done by air drag is -91.2 J

d)
from h = 0 m to 5 m

work dork by gravity from 0 to 5 m = m * g * h1

work dork by gravity from 0 to 5 m = 3 * 9.8 * 5

work dork by gravity from 0 to 5 m = 147 J

the work dork by gravity from 0 to 5 m is 147 J

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