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A 4.80 kilogram watermelon is dropped from rest from the roof of a 25m tall building. a.) calculate the work done by gravity on the watermelon. b.) Jut before it strikes the ground, what is the speed and KE? Assume no air resistance.

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Final answer:

The work done by gravity on the watermelon is 1176 Joules. Just before striking the ground, the watermelon's speed is 22.14 m/s, and its kinetic energy is 1176 Joules.

Step-by-step explanation:

Calculating Work Done by Gravity, Speed, and Kinetic Energy

When the watermelon is dropped from a height, gravity does work on it as it falls. The work done by gravity (W) can be calculated using the formula:

W = mgh

where m is the mass of the watermelon (4.80 kg), g is the acceleration due to gravity (9.80 m/s2), and h is the height (25 m).

Substituting the given values:

W = (4.80 kg)(9.80 m/s2)(25 m) = 1176 J (Joules)

The speed (v) just before the watermelon hits the ground can be found using the following kinematic equation for objects dropped from rest:

v = √(2gh)

Substituting in the values for g and h:

v = √(2(9.80 m/s2)(25 m)) = √(490 m2/s2) = 22.14 m/s

The kinetic energy (KE) of the watermelon just before impact can be calculated using the kinetic energy formula:

KE = 0.5mv2

Substituting the values:

KE = 0.5(4.80 kg)(22.14 m/s)2 = 1176 J

As expected, the kinetic energy just before impact is equal to the work done by gravity, assuming no air resistance.

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