Question

A 500 g object is dropped from a height of 2 meters. What is its kinetic energy just before it hits the ground?

Answer 1

Answer:

9.81 joules.

Step by step solved:

The kinetic energy (KE) of an object is given by the formula:

KE = 1/2 * m * v^2

where m is the mass of the object and v is its velocity.

When an object is dropped from rest, it gains speed as it falls due to the force of gravity. The velocity of a falling object can be calculated using the formula:

v = √(2gh)

where g is the acceleration due to gravity (9.81 m/s^2) and h is the height from which the object is dropped.

In this problem, the mass of the object is 500 g or 0.5 kg, and the height from which it is dropped is 2 meters. Using the formula for velocity, we get:

v = √(2gh) = √(2 x 9.81 m/s^2 x 2 m) = √(39.24) = 6.27 m/s

Now that we know the velocity of the object just before it hits the ground, we can calculate its kinetic energy using the formula:

KE = 1/2 * m * v^2 = 1/2 * 0.5 kg * (6.27 m/s)^2 = 9.81 J

Therefore, the kinetic energy of the 500 g object just before it hits the ground is 9.81 joules.

Answer 2

Answer: 9.8 J

Step-by-step explanation:

Since the gravitational potential energy of the object is mgh or mass*acceleration due to gravity*initial height, its
`U_(g)` is 9.8 J. Due to the Law of Conservation of Energy, its kinetic energy will also be 9.8 J. This can be seen in the equation
`KE_(i)+ PE_(i)= KE_(f) + PE_(f)`. Since there is no initial kinetic energy and no final potential energy, its initial potential energy is equal to its final kinetic energy.