Question

A double inclined plane as shown in the figure has fixed horizontal base and smooth faces with the same angle of inclination of $30^{\circ}$. A block of mass $300\, g$ is on one face and is connected by a cord passing over a frictionless pulley to a second block of mass $200\, g$ kept on another face. The acceleration with which the system of the blocks moves is w.... \% of acceleration due to gravity.

• A. 5
• B. 10
• C. 15
• D. 20

Answer 1

The Correct Option is B

 To solve the problem of determining the acceleration of a system of blocks on a double inclined plane, we begin by visualizing the forces acting on each block.

1. The angle of inclination for both planes is \(30^{\circ}\).
2. The mass of the block on one face (Block A) is \(300\,g = 0.3\,kg\).
3. The mass of the block on the other face (Block B) is \(200\,g = 0.2\,kg\).

Let's denote:

• \(g\) as the acceleration due to gravity, approximately \(9.8\,m/s^2\).
• \(a\) as the acceleration of the system.
• The tension in the cord as \(T\).

For Block A, on the inclined plane:

• The component of gravitational force acting down the plane is \(0.3g \sin(30^{\circ})\).

For Block B, on the inclined plane:

• The component of gravitational force acting down the plane is \(0.2g \sin(30^{\circ})\).

Since both surfaces are smooth (frictionless), we can write the equations of motion for each block:

1. For Block A: 

\[T - 0.3g \sin(30^{\circ}) = 0.3a\]

1. For Block B: 

\[0.2g \sin(30^{\circ}) - T = 0.2a\]

Solving these two equations simultaneously:

Add the two equations:

\[(T - 0.3g \sin(30^{\circ})) + (0.2g \sin(30^{\circ}) - T) = 0.3a + 0.2a\]

This simplifies to:

\[(0.2g - 0.3g) \sin(30^{\circ}) = 0.5a\]\[-0.1g \sin(30^{\circ}) = 0.5a\]

• Using \(\sin(30^{\circ}) = 0.5\), we have: 

\[-0.1g \times 0.5 = 0.5a\]\[-0.05g = 0.5a\]\[a = -0.1g = -0.1 \times 9.8 = -0.98 \, m/s^2\]

The negative sign indicates the direction of acceleration is opposite to the assumed direction. As asked, we find the percentage of \(a\) with respect to \(g\):

\[\frac{|a|}{g} \times 100 = \frac{0.98}{9.8} \times 100 = 10\%\]

Hence, the acceleration with which the system of blocks moves is 10% of acceleration due to gravity.