Question

A box of mass 5 kg is pulled by a cord, up along a frictionless plane inclined at 30° with the horizontal. The tension in the cord is 30 N. The acceleration of the box is (Take g = 10 m s^–2)

• A. 2 m s^–2
• B. Zero
• C. 0.1 m s^–2
• D. 1 m s^–2

Answer 1

The Correct Option is D

To find the acceleration of the box, analyze the forces acting on the box as it moves up the inclined plane. The key forces to consider are:

• Tension (\( T \)) in the cord, which is 30 N. 
• Gravitational force (\( mg \)), where \( m = 5 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
• Component of gravitational force parallel to the plane: This component can be calculated using \( mg \sin \theta \), where \( \theta = 30^\circ \).

Let's calculate the parallel component of weight:

\(mg \sin \theta = 5 \times 10 \times \sin 30^\circ = 50 \times 0.5 = 25 \, \text{N}\)

The net force (\( F_{\text{net}} \)) along the incline is given by:

\(F_{\text{net}} = T - mg \sin \theta\)

Substitute the values:

\(F_{\text{net}} = 30 \, \text{N} - 25 \, \text{N} = 5 \, \text{N}\)

This net force produces the acceleration of the box. According to Newton's second law:

\(F_{\text{net}} = ma\)

Therefore:

\(a = \frac{F_{\text{net}}}{m} = \frac{5}{5} = 1 \, \text{m/s}^2\)

The acceleration of the box is 1 m/s². Thus, the correct answer is 1 m/s².