A painter on a wooden plank drops a 1.50-kg can of paint from a height of 6 m. Find the following: (A) KE when the height is 4 m above the ground. (B) speed of the can 4 m above the ground. (C) speed of the can 1 m above the ground.

falls 2 meters

Ke = m g (2) = 1.5 * 9.8 * 2

(1/2) m v^2 = 1.5 * 9.8 * 2
or
v^2 = 9.8 * 4

falls 5 m from 6 to 1 meter
(1/2) m v^2 = m g (5)
v^2 = 9.8 * 10

in general falling distance h
v= sqrt (2 g h)

To find the answers to the given questions, we need to apply principles of gravitational potential energy, kinetic energy, and the conservation of mechanical energy.

(A) To find the kinetic energy (KE) when the height is 4 m above the ground, we can use the conservation of mechanical energy. The conservation of mechanical energy states that the total mechanical energy of the system remains constant as long as no external forces act on it.

The total mechanical energy (E) of the system is the sum of the potential energy (PE) and the kinetic energy (KE):

E = PE + KE

As the can is dropped from a height of 6 m, we can assume that the potential energy at that point is converted entirely into kinetic energy.

Initially, when the can is at a height of 6 m, its potential energy is given by:

PE1 = mgh1

where m is the mass of the can (1.50 kg), g is the acceleration due to gravity (9.8 m/s^2), and h1 is the initial height (6 m).

So, PE1 = (1.50 kg) * (9.8 m/s^2) * (6 m) = 88.2 J

At a height of 4 m, the potential energy becomes:

PE2 = mgh2

where h2 is the final height (4 m).

So, PE2 = (1.50 kg) * (9.8 m/s^2) * (4 m) = 58.8 J

Since mechanical energy is conserved, the final kinetic energy is:

KE = E - PE2 = PE1 - PE2 = 88.2 J - 58.8 J = 29.4 J

Therefore, the kinetic energy of the can when it is at a height of 4 m above the ground is 29.4 J.

(B) To find the speed of the can when it is 4 m above the ground, we can use the equation for kinetic energy:

KE = (1/2)mv^2

where v is the velocity or speed of the can.

Rearranging the equation, we have:

v^2 = (2KE)/m

Plugging in the values, we get:

v^2 = (2 * 29.4 J) / 1.50 kg = 39.2 m^2/s^2

Taking the square root of both sides, we find:

v = √(39.2 m^2/s^2) = 6.26 m/s

Therefore, the speed of the can when it is 4 m above the ground is approximately 6.26 m/s.

(C) To find the speed of the can when it is 1 m above the ground, we can use the same equation:

v^2 = (2KE)/m

The kinetic energy at this point is still equal to 29.4 J (as found in part A), and the mass of the can remains 1.50 kg.

Plugging in the values, we get:

v^2 = (2 * 29.4 J) / 1.50 kg = 39.2 m^2/s^2

Taking the square root of both sides, we find:

v = √(39.2 m^2/s^2) = 6.26 m/s

Therefore, the speed of the can when it is 1 m above the ground is also approximately 6.26 m/s.

Thank you

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