A painter on a scaffold drops a 1.25 kg can of paint from a height of 5.00 m. a) What is the kinetic energy of the can when the can is at a height of 3.00 m?
b) With what speed will the can hit the ground? (Neglect air resistance)

(a) The kinetic energy of the can when the can is at a height of 3.00 m.

(b) The speed with which the can will hit the ground is 9.9 m/s

Explanation:

(a) Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and at rest, until it reaches a given speed.

Kinetic energy is represented by the following formula:

Ec = ½ mv²

Where Ec is kinetic energy, which is measured in Joules (J), m is mass measured in kilograms (kg), and v is velocity measured in meters over seconds (m/s).

On the other hand, potential energy is the energy that measures the ability of a system to perform work based on its position. In other words, this is the energy that a body has at a certain height above the ground.

Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the force of gravity. Then for an object with mass m, at height h, the expression applied to the gravitational energy of the object is:

Ep = m g h

Where Ep is the potential energy in joules (J), m is the mass in kilograms (kg) is h the height in meters (m) and g is the acceleration of fall in m/s² (approximately 9.8 m/s²)

Energy is neither created nor destroyed, but only transforms from one state to another. This principle also extends to mechanical energy, so that in an isolated system, the sum of kinetic and potential energies between two instants of time remains constant:

½ m*v1² + m*g*h1=  ½ m*v2² + m*g*h2

 where v1 and h1 are the initial speed and height and v2 and h2 the final speed and height.

In this case:

Replacing:

½ m*0² + m*g*5=  ½ m*v2² + m*g*3

Solving:

m*g*5= ½ m*v2² + m*g*3

The mass m appears in all terms, being able to simplify:

g*5=  ½ v2² + g*3

Solving for v2:

being g= 9.8 m/s²:

v2= 6.26 m/s

The speed at 3 meters height is 6.26 m/s

Then the kinetic energy is calculated as:

Ec = ½ mv²

Ec = ½ 1.25 kg* (6.26 m/s)²

Ec= 24.49 J

The kinetic energy of the can when the can is at a height of 3.00 m.

(b) You know:

½ m*v1² + m*g*h1=  ½ m*v2² + m*g*h2

 where v1 and h1 are the initial speed and height and v2 and h2 the final speed and height.

In this case:

Replacing:

½ m*0² + m*g*5=  ½ m*v2² + m*g*0

Solving:

m*g*5= ½ m*v2²

g*5=  ½ v2²

being g= 9.8 m/s²:

v2= 9.9 m/s

The speed with which the can will hit the ground is 9.9 m/s

(75°f − 32) × 5/9 = 23.889°c

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