Question

A machine with efficiency \( \frac{2}{3} \) used 12 J of energy in lifting a 2 kg block through a certain height and it is allowed to fall through the same. The velocity while it reaches the ground is:

• A. \( \sqrt{2} \, \text{ms}^{-1} \)
• B. \( 2 \, \text{ms}^{-1} \)
• C. \( 2 \sqrt{2} \, \text{ms}^{-1} \)
• D. \( 0.2 \, \text{ms}^{-1} \)

Hint:

When dealing with energy problems, use the conservation of energy principle, where potential energy is converted to kinetic energy. Don't forget to account for efficiency in the system.

Answer 1

The Correct Option is C

Step 1: Analyze the energy transfer Given: 
- The efficiency of the machine is \( \frac{2}{3} \). 
- The total energy used is 12 J. 
- The mass of the block is 2 kg. 
The total energy used is \( 12 \, \text{J} \), but the machine only uses \( \frac{2}{3} \) of that energy to lift the block, so the effective energy that goes into raising the block is: \[ \text{Effective Energy} = \frac{2}{3} \times 12 = 8 \, \text{J} \] 
Step 2: Use the energy equation to calculate velocity The potential energy at the height to which the block is raised is equal to the effective energy used by the machine. The potential energy is given by: \[ PE = mgh \] where: 
- \( m = 2 \, \text{kg} \) (mass of the block), 
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), 
- \( h \) is the height the block is lifted. Using the relation \( PE = mgh = 8 \, \text{J} \), we can solve for \( h \): \[ 8 = 2 \times 9.8 \times h \] \[ h = \frac{8}{2 \times 9.8} = \frac{8}{19.6} = 0.408 \, \text{m} \] Now, the block falls through the same height \( h \), and its potential energy is converted into kinetic energy when it reaches the ground. The kinetic energy at the ground is: \[ KE = \frac{1}{2} m v^2 \] Equating the potential energy and kinetic energy: \[ 8 = \frac{1}{2} \times 2 \times v^2 \] Solving for \( v \): \[ v^2 = \frac{8}{1} = 8 \quad \Rightarrow \quad v = \sqrt{8} = 2 \sqrt{2} \, \text{ms}^{-1} \] Thus, the velocity of the block when it reaches the ground is \( 2 \sqrt{2} \, \text{ms}^{-1} \).

Answer 2

Given:
Efficiency, \( \eta = \frac{2}{3} \)
Energy used by machine, \( E = 12 \, \text{J} \)
Mass of block, \( m = 2 \, \text{kg} \)

Step 1: Calculate the useful work done (i.e., the potential energy gained by the block):
\[ \text{Useful work} = \eta \times E = \frac{2}{3} \times 12 = 8 \, \text{J} \]

Step 2: Calculate the height \( h \) through which the block is lifted:
\[ mgh = 8 \implies h = \frac{8}{mg} = \frac{8}{2 \times 9.8} \approx \frac{8}{19.6} = 0.408 \, \text{m} \]

Step 3: When the block falls from height \( h \), potential energy converts to kinetic energy:
\[ mgh = \frac{1}{2} m v^2 \implies v = \sqrt{2gh} \]

Step 4: Substitute the value of \( h \):
\[ v = \sqrt{2 \times 9.8 \times 0.408} \approx \sqrt{8} = 2 \sqrt{2} \, \text{m/s} \]

Therefore, the velocity of the block when it reaches the ground is:
\[ \boxed{2 \sqrt{2} \, \text{ms}^{-1}} \]