Question

A 2.5-kg brick falls to the ground from a 3-m-high roof. What is the approximate kinetic energy of the brick just before it touches the ground?

Answer 1

Question:

A 2.5-kg brick falls to the ground from a 3-m-high roof. What is the approximate kinetic energy of the brick just before it touches the ground?

a) 75J

b) 12J

c) 38J

d) 11J

Answer:

Kinetic Energy of the brick just before it touches ground = 73.5 Joules, so according to the question, approximately correct options is 75 Joules.

Step-by-step explanation:

Mass of brick = m = 2.5 kg

Height of roof = 3 m

Let us assume that the final velocity of the brick just before it touches the ground is v.

Kinetic energy of a body is given by: KE =
`(1)/(2) * m * v^2`

We need to calculate v before calculating KE.

By the Equation
`v^2 = u^2 + 2 * a * s`, where u is initial velocity, v is final velocity, a is acceleration, and s is distance traveled, we can calculate the required variable.

In this case, s = Height of roof, u = 0 because the brick is not thrown, but just falls, and a = gravitational acceleration = g = 9.8
`(m)/(s^2)`.

So
`v^2 = 0 + 2 * 9.8 * 3` = 58.8

So
`v = √(58.8)` = 7.66
`(m)/(s)`

So now we can calculate Kinetic Energy =
`(1)/(2) * m * v^2` =
`(1)/(2) * 2.5 * 58.8` = 73.5
`kg * (m^2)/(s^2)`

Also we need to convert KE into Joules.

1
`kg * (m^2)/(s^2)` = 1 Joule

So Kinetic Energy = 73.5 Joules.