Question

A body is made to move up along an inclined plane of inclination \(30^\circ\) and the coefficient of friction is 0.5, then its retardation is:

• A. \( \frac{2 + \sqrt{3}}{4} g \)
• B. \( \frac{2 - \sqrt{3}}{4} g \)
• C. \( \frac{2 - \sqrt{3}}{2} g \)
• D. \( \frac{2 + \sqrt{3}}{2} g \)

Hint:

When solving for acceleration or retardation on an inclined plane, use the formula \( a = g \left( \sin \theta + \mu \cos \theta \right) \) and substitute the values of \( \theta \) and \( \mu \).

Answer 1

The Correct Option is A

We are given an inclined plane of inclination \(30^\circ\) and the coefficient of friction \( \mu = 0.5 \). The retardation \( a \) is the acceleration due to gravity \( g \) modified by the forces acting on the body. For a body moving up an inclined plane, the retardation is given by: \[ a = g \left( \sin \theta + \mu \cos \theta \right) \] where \( \theta = 30^\circ \) and \( \mu = 0.5 \). \[ a = g \left( \sin 30^\circ + 0.5 \cos 30^\circ \right) \] Substituting the known values \( \sin 30^\circ = \frac{1}{2} \) and \( \cos 30^\circ = \frac{\sqrt{3}}{2} \), we get: \[ a = g \left( \frac{1}{2} + 0.5 \times \frac{\sqrt{3}}{2} \right) \] \[ a = g \left( \frac{1}{2} + \frac{\sqrt{3}}{4} \right) \] \[ a = g \left( \frac{2 + \sqrt{3}}{4} \right) \] Thus, the retardation is \( \frac{2 + \sqrt{3}}{4} g \).

Answer 2

Step 1: Given values
- Inclination angle \( \theta = 30^\circ \)
- Coefficient of friction \( \mu = 0.5 \)

Step 2: Forces involved
- Gravitational component down the incline: \( mg \sin\theta \)
- Frictional force (also acting down since body is moving up): \( \mu mg \cos\theta \)
- Net retardation \( a = g(\sin\theta + \mu \cos\theta) \)

Step 3: Substitute known values
\[ a = g(\sin 30^\circ + 0.5 \cos 30^\circ) = g\left(\frac{1}{2} + 0.5 \times \frac{\sqrt{3}}{2}\right) = g\left(\frac{1 + \sqrt{3}}{2} \times \frac{1}{2} \right) = \frac{2 + \sqrt{3}}{4} g \]

Final Answer:
\[ \boxed{\frac{2 + \sqrt{3}}{4} g} \]