A 4.20-kg watermelon is dropped from rest from the roof of a 28.0-m -tall building and feels no appreciable air resistance. Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. Just before it strikes the ground, what is the watermelon's kinetic energy? Just before it strikes the ground, what is the watermelon's speed? Would the answer in part A be different if there were appreciable air resistance? Would the answer in part B be different if there were appreciable air resistance? Would the answer in part C be different if there were appreciable air resistance?

Well, well, well, the watermelon sure knows how to make an entrance! Let's crunch some numbers and have some fun, shall we?

First, let's calculate the work done by gravity during the watermelon's fall. We can use the equation:

Work = Force * Distance * cosine(angle)

Since the watermelon is falling vertically, the angle is 0 degrees, and the cosine of 0 degrees is 1. So the equation simplifies to:

Work = Force * Distance

The force acting on the watermelon is its weight, given by:

Force = mass * gravity

Plugging in the values, we get:

Force = 4.20 kg * 9.8 m/s²

Calculating the force gives us a value of 41.16 N (approximately).

Now, let's calculate the work:

Work = 41.16 N * 28.0 m

And the answer is... 1152.48 Joules (approximately).

Now, let's move on to the kinetic energy just before the watermelon strikes the ground. The formula for kinetic energy is:

Kinetic Energy = (1/2) * mass * velocity²

Since the watermelon starts from rest, the initial velocity is 0.

So, the kinetic energy just before it strikes the ground is:

Kinetic Energy = (1/2) * 4.20 kg * velocity²

But, hold your watermelon! To find the velocity just before impact, we need to use an energy conservation approach. The potential energy it had when it was dropped from the roof is fully converted into kinetic energy just before it hits the ground.

Potential Energy = Kinetic Energy

m * g * h = (1/2) * m * v²

Canceling out the mass, we find:

g * h = (1/2) * v²

Substituting the values, we get:

9.8 m/s² * 28.0 m = (1/2) * v²

Now we can solve for the velocity:

v² = (9.8 m/s² * 28.0 m) / (1/2)

And voila! After doing the math, the velocity just before impact is approximately 19.84 m/s.

Now, let's address the question about air resistance. If there were appreciable air resistance, the answer would indeed be different. Air resistance would counteract the motion of the watermelon, slowing it down. As a result, it would require less work to be done by gravity, and the final velocity of the watermelon would be lower.

So, the answer in part A would be different if there were appreciable air resistance, resulting in less work being done by gravity.

Similarly, in part B, with air resistance, the watermelon's kinetic energy just before striking the ground would be lower, as it would have lost some of its energy due to air resistance.

Finally, in part C, if there were appreciable air resistance, the watermelon's speed just before striking the ground would also be lower, again due to the counteracting force of air resistance.

Hope that clears everything up, and remember to always keep your watermelons well-secured before dropping them from tall buildings!

To calculate the work done by gravity on the watermelon, we need to use the equation:

Work = Force x Distance x cos(θ)

In this case, the force is the weight of the watermelon due to gravity, which can be calculated using:

Weight = mass x gravitational acceleration

The distance is the height of the building, which is given as 28.0 m.

The angle (θ) between the force and the displacement is 0 degrees because the gravitational force acts vertically downward, and the displacement is also vertical.

First, let's calculate the weight of the watermelon:

Weight = mass x gravitational acceleration
Weight = 4.20 kg x 9.8 m/s^2
Weight = 41.16 N

Now, let's calculate the work done by gravity:

Work = Force x Distance x cos(θ)
Work = 41.16 N x 28.0 m x cos(0)
Work = 1154.08 J

Therefore, the work done by gravity on the watermelon is 1154.08 Joules.

Just before the watermelon strikes the ground, all of its potential energy due to its height has been converted into kinetic energy. So, at that point, the watermelon's kinetic energy is equal to the potential energy it had at the roof of the building.

Potential Energy = mass x gravitational acceleration x height

Potential Energy = 4.20 kg x 9.8 m/s^2 x 28.0 m
Potential Energy = 1169.76 J

Therefore, the watermelon's kinetic energy just before it strikes the ground is 1169.76 Joules.

To find the watermelon's speed just before it strikes the ground, we can use the equation:

Kinetic Energy = (1/2) x mass x velocity^2

1169.76 J = (1/2) x 4.20 kg x velocity^2
2339.52 J = 2.10 kg x velocity^2

Dividing both sides by 2.10 kg and taking the square root:

velocity^2 = (2339.52 J) / (2.10 kg)
velocity^2 = 1114.06 m^2/s^2

velocity = √(1114.06 m^2/s^2)
velocity ≈ 33.4 m/s

Therefore, just before the watermelon strikes the ground, its speed is approximately 33.4 m/s.

If there were appreciable air resistance, both the work done by gravity and the watermelon's kinetic energy would be different. Air resistance would oppose the motion of the watermelon, resulting in some of its energy being lost to air. The work done by gravity would be lower, as part of the potential energy would be converted into work against air resistance. Similarly, the watermelon's kinetic energy and speed would also be lower due to the energy loss caused by air resistance. The actual values would depend on the magnitude of the air resistance forces present.

(a)W(gravity) = mgh

(b) mgh =mv²/2
v=sqrt(2gh)
(c) the same
(d)
mgh = W(fr) + mv1²/2
v1 <v

(d) Just before it strikes the ground, what is the watermelon's speed?

Vterminal = sqrt(2*acceleration*displacement) = sqrt(2*9.8 m/s^2*20 m) = 19.80 m/s

a) W=m*g*h

W=(4.20kg)(28.0m)(9.8m/s^2)=1152J
b)K=1/2*m*v^2 & v=sqrt2*g*h
v=sqrt(2)(9.8m/s^2)(28.0m)=23.4 m/s
K=(1/2)(28.0m)(23.4)^2=7665.8J
c)23.4m/s
d)no.
e)yes. Air resistance would do negative work
f)yes. Air resistance would do negative work