Question

A body of mass 250 g falls from a tower of height
20 m. What will be its kinetic energy when it has
fallen (1/4)th of the height of tower? (Take, g =
10 m/s2)​

Answer 1

Answer:

12.5 J

Explanation:

Given:

• Acceleration (a) = g = 10 m/s²

• Initial velocity (u) = 0 m/s

• Mass of the body (m) = 250 g = 1/4 kg

• Height of the tower = 20 m

So, 1/4 height of the tower = 20/4 = 5 m = Displacement (s)

To find:

Kinetic Energy when it has fallen 1/4th of the height of the tower. (K.E)

Method to find:

First, we need to find its final velocity (v). We can do this by using the 3rd formula of motion.

v² = u² + 2as

⇒ v² = 0 + (2*10*5)

⇒ v² = 100

⇒ v = √100

⇒ v = 10 m/s

Now, we can find the kinetic energy of the body.

K.E = 1/2mv²

⇒ K.E = 1/2 * 1/4 * 10 * 10

⇒ K.E = 100/8

⇒ K.E = 12.5 Joules

Hope it helps!

Answer 2

Given:-

→ Mass of the body = 250 g

→ Height of the tower = 20 m

→ Value of 'g' = 10 m/s²

To find:-

→ K.E. of the body when it has fallen

(1/4)th of the height of the tower.

Solution:-

Firstly, let's convert mass of the body from g to kg.

=> 1 g = 0.001 kg

=> 250 g = 250(0.001)

=> 0.25 kg

________________________________

Required height :-

=> 1/4 × 20

=> 20/4

=> 5m

Now, let's calculate the velocity of the body when it has fallen 5m, with the help of 3rd equation of motion :-

=> v² = u² + 2gh

=> v² = 0 + 2(10)(5)

=> v² = 100

=> v = √100

=> v = 10 m/s

We know that :-

K.E. = ½mv²

Where :-

• K.E. is kinetic energy of the body.

• m is mass of the body.

• v is the velocity of the body.

Substituting values, we get :-

=> K.E. = ½ × 0.25 × 10 × 10

=> K.E. = 0.25 × 5 × 10

=> K.E. = 12.5 J

Thus, kinetic energy of the body is 12.5J .