A mass of 0.5 kg is just able to slide down the slope of an inclined rough surface when the angle of inclination is `60^(@)`. The minimum force necessary to pull the mass up the incline along the line of greatest slope is : (Take `g=10m//s^(2)`)

Options

A. 20 N
B. 9 N
C. 100 N
D. 1 N

Answer

To solve the problem step by step, we will analyze the forces acting on the mass on the inclined plane and calculate the minimum force necessary to pull the mass up the incline.

### Step 1: Identify the forces acting on the mass
- The weight of the mass (W) acting downwards: \( W = mg \)
- The normal force (N) acting perpendicular to the incline
- The frictional force (F_f) acting up the incline when sliding down
- The component of the weight acting down the incline: \( W_{\text{down}} = mg \sin \theta \)
- The component of the weight acting perpendicular to the incline: \( W_{\text{perpendicular}} = mg \cos \theta \)

### Step 2: Calculate the weight of the mass
Given:
- Mass (m) = 0.5 kg
- Acceleration due to gravity (g) = 10 m/s²

Calculating the weight:
\[ W = mg = 0.5 \times 10 = 5 \, \text{N} \]

### Step 3: Calculate the components of the weight
Using the angle of inclination \( \theta = 60^\circ \):
- \( W_{\text{down}} = mg \sin(60^\circ) = 5 \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \)
- \( W_{\text{perpendicular}} = mg \cos(60^\circ) = 5 \cos(60^\circ) = 5 \times \frac{1}{2} = 2.5 \, \text{N} \)

### Step 4: Calculate the frictional force
The frictional force (F_f) when the mass is just about to slide down can be expressed as:
\[ F_f = \mu N \]
Where \( N = W_{\text{perpendicular}} = 2.5 \, \text{N} \).

To find the coefficient of friction \( \mu \):
Since the mass can just slide down at \( \theta = 60^\circ \):
\[ F_f = mg \sin(60^\circ) \]
Thus:
\[ \mu N = mg \sin(60^\circ) \]
Substituting \( N \):
\[ \mu (2.5) = \frac{5\sqrt{3}}{2} \]
Solving for \( \mu \):
\[ \mu = \frac{5\sqrt{3}/2}{2.5} = \sqrt{3} \]

### Step 5: Calculate the minimum force to pull the mass up the incline
To pull the mass up the incline, the applied force \( F \) must overcome both the frictional force and the component of the weight acting down the incline:
\[ F + F_f = mg \sin(60^\circ) \]
Substituting \( F_f = \mu N = \sqrt{3} \times 2.5 \):
\[ F + \sqrt{3} \times 2.5 = \frac{5\sqrt{3}}{2} \]
Rearranging gives:
\[ F = \frac{5\sqrt{3}}{2} - \sqrt{3} \times 2.5 \]
Factoring out \( \sqrt{3} \):
\[ F = \sqrt{3} \left( \frac{5}{2} - 2.5 \right) = \sqrt{3} \left( \frac{5 - 5}{2} \right) = 0 \]
This indicates that we need to consider the total force needed to overcome both the weight and friction when pulling up the incline:
\[ F = \mu mg \cos(60^\circ) + mg \sin(60^\circ) \]
Substituting the values:
\[ F = \sqrt{3} \times 2.5 + \frac{5\sqrt{3}}{2} \]
Calculating:
\[ F = \sqrt{3} \times 2.5 + \frac{5\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} + \frac{5\sqrt{3}}{2} = 5\sqrt{3} \]
Now substituting \( \sqrt{3} \approx 1.73 \):
\[ F \approx 5 \times 1.73 = 8.65 \, \text{N} \]
Thus, the minimum force necessary to pull the mass up the incline is approximately 9 N.

### Final Answer
The minimum force necessary to pull the mass up the incline is approximately **9 N**.