Question

A 75 kg box is dropped from the top of a tower. The height of the tower is 35 m. Calculate (i) the initial
potential energy of the box, (ii) its potential energy 15 m above the ground, (iii) the maximum value of
its kinetic energy and (iv) its kinetic energy 20 m below the top of the tower. (g = 9.8 m/s2)
(g = 9.8 m/s2)​

Answer 1

Given :-

• A 75 kg box is dropped from the top of a tower of height 35 m.
• Acceleration due to gravity, g = 9.8 m/s²

To Find :-

• Initial potential energy.
• Potential energy 15 m above the ground.
• Maximum value of kinetic energy
• Kinetic energy 20 m below the top of the tower.

Solution :-

First, Let us find the Initial potential energy at the top of the tower,

• ⇒ PE = mgh
• ⇒ PE = 75 × 9.8 × 35
• ⇒ PE = 25,725 J

_______________________

Here, We have

• Height, h = 15 m above the ground.
• g = 9.8 m/s²
• mass = 75 kg

• ⇒ PE = mgh
• ⇒ PE = 75 × 9.8 × 15
• ⇒ PE = 11,025 J

_______________________

• Maximum value of the kinetic energy will be at the very bottom of the tower ( touching the ground ) where all the potential energy will be converted into Kinetic energy (According to law of conservation of energy)

So,

⇒ Maximum Kinetic energy = PE at the top of tower

⇒ Maximum Kinetic energy = 25,725 J

_______________________

• Here, we have to find the kinetic energy 20 m below the top of the tower.

Let's find the potential energy at this point, we have

• Height, h = 35 - 20 = 15 m
• g = 9.8 m/s²
• mass = 75 kg

So,

• ⇒ PE = mgh
• ⇒ PE = 75 × 9.8 × 15
• ⇒ PE = 11,025 J

• But, The total kinetic energy should be 25,725 J including the kinetic energy, according to the law of conservation of energy,

• ⇒ Total energy = PE + KE
• ⇒ 25,725 = 11,025 + KE
• ⇒ KE = 25,725 - 11,025
• ⇒ KE = 14,700 J