A block of mass 5 kg is placed on a rough inclined surface as shown in the figure.
If \(\vec{F_1}\) is the force required to just move the block up the inclined plane and \(\vec{F_2}\) is the force required to just prevent the block from sliding down, then the value of \(|\vec{F_1}| - |\vec{F_2}|\) is: [Use \(g = 10 \, \text{m/s}^2\)]
If \(\vec{F_1}\) is the force required to just move the block up the inclined plane and \(\vec{F_2}\) is the force required to just prevent the block from sliding down, then the value of \(|\vec{F_1}| - |\vec{F_2}|\) is: [Use \(g = 10 \, \text{m/s}^2\)]
- \(25 \sqrt{3} \, \text{N}\)
- \(5\sqrt{3} \, \text{N}\)
- \(\frac{5 \sqrt{3}}{2} \, \text{N}\)
- \(10 \, \text{N}\)
The Correct Option is B
Approach Solution - 1
To solve this problem, we need to calculate the forces \(|\vec{F_1}|\) and \(|\vec{F_2}|\), and then find the difference \(|\vec{F_1}| - |\vec{F_2}|\).
Given:
- Mass of block (\(m\)) = 5 kg
- Inclination angle (\(\theta\)) = \(30^\circ\)
- Coefficient of friction (\(\mu\)) = 0.1
- Acceleration due to gravity (\(g\)) = 10 m/s²
The forces on the block parallel and perpendicular to the inclined plane are:
\(F_{\text{gravity parallel}} = mg \sin(\theta)\)
\(F_{\text{gravity perpendicular}} = mg \cos(\theta)\)
The limiting friction is given by:
\(F_{\text{friction}} = \mu \cdot F_{\text{gravity perpendicular}} = \mu \cdot mg \cos(\theta)\)
Calculating \(|\vec{F_1}|\) (force to move block up):
To move the block up, apply enough force to overcome both gravity and friction:
\(|\vec{F_1}| = F_{\text{gravity parallel}} + F_{\text{friction}}\)
\(= mg \sin(\theta) + \mu mg \cos(\theta)\)
Substituting values:
\(|\vec{F_1}| = 5 \cdot 10 \cdot \sin(30^\circ) + 0.1 \cdot 5 \cdot 10 \cdot \cos(30^\circ)\)
\(= 50 \cdot 0.5 + 0.1 \cdot 50 \cdot \frac{\sqrt{3}}{2}\)
\(= 25 + 2.5\sqrt{3}\)
Calculating \(|\vec{F_2}|\) (force to prevent sliding down):
To prevent sliding down, counteract gravity by friction:
\(|\vec{F_2}| = F_{\text{gravity parallel}} - F_{\text{friction}}\)
\(= mg \sin(\theta) - \mu mg \cos(\theta)\)
Substituting values:
\(|\vec{F_2}| = 5 \cdot 10 \cdot \sin(30^\circ) - 0.1 \cdot 5 \cdot 10 \cdot \cos(30^\circ)\)
\(= 25 - 2.5\sqrt{3}\)
Calculating the difference:
\(|\vec{F_1}| - |\vec{F_2}| = (25 + 2.5\sqrt{3}) - (25 - 2.5\sqrt{3})\)
\(= 5\sqrt{3} \, \text{N}\)
Thus, the correct answer is \(5\sqrt{3} \, \text{N}\).
Approach Solution -2
The block experiences frictional force due to the roughness of the inclined surface. The kinetic friction force \( f_k \) is given by:
\[ f_k = \mu mg \cos \theta, \]where \( \mu = 0.1 \) is the coefficient of friction, \( m = 5 \, \text{kg} \), and \( \theta = 30^\circ \).
Calculating \( f_k \):
\[ f_k = 0.1 \times 5 \times 10 \times \cos 30^\circ = 0.1 \times 50 \times \frac{\sqrt{3}}{2} = 2.5 \sqrt{3} \, \text{N}. \]To move the block up the incline, the force \( F_1 \) must overcome both the component of gravitational force along the incline and the frictional force. Therefore:
\[ F_1 = mg \sin \theta + f_k. \]Substitute the values:
\[ F_1 = 5 \times 10 \times \sin 30^\circ + 2.5 \sqrt{3} = 25 + 2.5 \sqrt{3} \, \text{N}. \]To prevent the block from sliding down, the force \( F_2 \) must balance the component of gravitational force along the incline, reduced by the frictional force. Thus:
\[ F_2 = mg \sin \theta - f_k. \]Substitute the values:
\[ F_2 = 25 - 2.5 \sqrt{3} \, \text{N}. \]The difference \( |F_1 - F_2| \) is:
\[ |F_1 - F_2| = |(25 + 2.5 \sqrt{3}) - (25 - 2.5 \sqrt{3})| = 5 \sqrt{3} \, \text{N}. \]Thus, the answer is:
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