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.

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 = $\frac{1}{2} \times m \times v^2$

We need to calculate v before calculating KE.

By the Equation $v^2 = u^2 + 2 \times a \times s$, 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 $\frac{m}{s^2}$.

So $v^2 = 0 + 2 \times 9.8 \times 3$ = 58.8

So $v = \sqrt{58.8}$ = 7.66 $\frac{m}{s}$

So now we can calculate Kinetic Energy = $\frac{1}{2} \times m \times v^2$ = $\frac{1}{2} \times 2.5 \times 58.8$ = 73.5 $kg \times \frac{m^2}{s^2}$

Also we need to convert KE into Joules.

1 $kg \times \frac{m^2}{s^2}$ = 1 Joule

So Kinetic Energy = 73.5 Joules.