Question

In the given figure, the block of mass m is dropped from the point ‘A’. The expression for kinetic energy of block when it reaches point ‘B’ is

Fig. 

• A. \(\frac{1}{2} mgy^2_{0}\)
• B. \(\frac{1}{2} mgy^2\)
• C. mg(y - y₀)
• D. mgy₀

Answer 1

The Correct Option is D

To determine the kinetic energy of the block when it reaches point ‘B’, we can use the principle of conservation of energy. Initially, at point ‘A’, the block has potential energy but no kinetic energy. As it falls to point ‘B’, this potential energy is converted into kinetic energy.

Step 1: Calculate Initial Potential Energy at Point ‘A’

The potential energy \( PE \) at height \( y_0 \) is given by:

PE = mgy_0

where:

• m is the mass of the block
• g is the acceleration due to gravity
• y_0 is the initial height

Step 2: Consider Energy Conservation

When the block reaches point ‘B’, its potential energy is zero (assuming the height of point ‘B’ is zero). According to the conservation of energy, the initial potential energy is converted entirely into kinetic energy at point ‘B’.

Step 3: Determine Kinetic Energy at Point ‘B’

The kinetic energy \( KE \) at point ‘B’ is equal to the initial potential energy:

KE = mgy_0

This implies the kinetic energy of the block when it reaches point ‘B’ is \( mgy_0 \).

Conclusion:

The correct expression for the kinetic energy of the block at point ‘B’ is mgy₀. This corresponds to option 4.

Fig.