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 = [tex]\frac{1}{2} \times m \times v^2[/tex]
We need to calculate v before calculating KE.
By the Equation [tex]v^2 = u^2 + 2 \times a \times s[/tex], 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 [tex]\frac{m}{s^2}[/tex].
So [tex]v^2 = 0 + 2 \times 9.8 \times 3[/tex] = 58.8
So [tex]v = \sqrt{58.8}[/tex] = 7.66 [tex]\frac{m}{s}[/tex]
So now we can calculate Kinetic Energy = [tex]\frac{1}{2} \times m \times v^2[/tex] = [tex] \frac{1}{2} \times 2.5 \times 58.8[/tex] = 73.5 [tex]kg \times \frac{m^2}{s^2}[/tex]
Also we need to convert KE into Joules.
1 [tex]kg \times \frac{m^2}{s^2}[/tex] = 1 Joule
So Kinetic Energy = 73.5 Joules.
