Question

A body of mass 0.6kg is thrown vertically upward from the ground with a speed of 20 m/s. Calculate its Potential energy at the maximum height reached

Answer 1

Given :

▪ Mass of body = 0.6kg

▪ Initial velocity = 20mps

To Find :

▪ Potential energy of body at maximum height.

Solution :

→ First we have to find out maximum height attained by the body after that we can calculate potential energy of body at maximum height.

→ Since, acceleration due to gravity is constant throughout the motion, we can easily apply equation of kinematics to solve this type of questions.

→ Third equation of kinematics is given by

☞ v^2 - u^2 = 2gH

• v denotes final velocity
• u denotes initial velocity
• g denotes acceleration due to gravity
• H denotes height

✒ Final velocity = zero (i.e. highest point)

✒ Sign of g is negative due to opposite direction.

✏ v^2 - u^2 = 2gH

✏ (0)^2 - u^2 = 2(-g)H

✏ H = u^2/(2g)

✏ H = (20×20)/(2×10)

✏ H = 20m

✴ Potential energy at max. height :

☞ U = mgH

☞ U = 0.6×10×20

☞ U = 6×20

☞ U = 120J

Additional information :

• Potential energy is a scalar quantity.
• It has only magnitude.

Answer 2

Given :

• Initial velocity (u) = 20 m/s
• Mass of body (m) = 0.6 kg
• Final velocity (v) = 0 m/s

To Find :

• Potential Energy when maximum height is reached

Solution :

We have to find the maximum height attained by the body, after that putting the values in Kinetic Energy formula, we can find value of Kinetic Energy :

$\underbrace{\sf{Maximum \: height}}$

Use 3rd equation of Kinematics

$\implies \sf{v^2 \: - \: u^2 \: = \: 2gh} \\ \\ \implies \sf{0^2 \: - \: 20^2 \: = \: 2 \: \times \: -10 \: \times \: h} \\ \\ \implies \sf{-400 \: = \: -20h} \\ \\ \implies \sf{h \: = \: \dfrac{400}{20}} \\ \\ \implies \sf{h \: = \: 20}$

$\therefore$ Maximum height reached is 20 m

__________________________________

$\underbrace{\sf{Potential \: Energy}}$

Use formula for Potential Energy

$\implies \sf{P.E \: = \: mgh} \\ \\ \implies \sf{P.E \: = \: 0.6 \: \times \: 10 \: \times \: 20} \\ \\ \implies \sf{P.E \: = \: 0.6 \: \times \: 200} \\ \\ \implies \sf{P.E \: = \: 120}$

$\therefore$ Potential energy at maximum height is 120 J