Question

If a 6 kg block is released from rest as shown in the figure, then find the velocity of the 6 kg block just before hitting the ground.

• A. 6.2 m/sec
• B. 7.74 m/sec
• C. 4.7 m/sec
• D. 3.87 m/sec

Hint:

Use the principle of conservation of energy for problems where only conservative forces (like gravity) are acting. Potential energy is converted into kinetic energy, and the total energy remains constant.

Answer 1

The Correct Option is B

Given the equation: \[ +6 \times 10 \times 6 - 2 \times 10 \times 6 \] Step 1: Simplifying the expression:
First, expand the terms: \[ = \frac{1}{2} \times 6 V^2 + \frac{1}{2} \times 2 V^2 \]
Step 2: Combine the terms:
Simplifying the right-hand side: \[ 360 - 120 = 4V^2 \]
Step 3: Solve for \( V^2 \):
\[ 4V^2 = 240 \]
Step 4: Solve for \( V \):
\[ V^2 = \frac{240}{4} = 60 \] Taking the square root of both sides: \[ V = \sqrt{60} \]
Step 5: Final simplification:
\[ V = 2 \sqrt{15} \] Thus, the final value of \( V \) is: \[ V = 2 \sqrt{15} = 7.74 \]