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kotykmax [81]
3 years ago
10

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 touc

hes the ground?
Physics
1 answer:
Marat540 [252]3 years ago
3 0
<h3><u>Question:</u></h3>

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

<h3><u>Answer:</u></h3>

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

<h3><u>Explanation:</u></h3>

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.

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