Question

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 $g = 10 \, \text{m/s}^2$]

• A. $25 \sqrt{3} \, \text{N}$
• B. $5\sqrt{3} \, \text{N}$
• C. $\frac{5 \sqrt{3}}{2} \, \text{N}$
• D. $10 \, \text{N}$

Answer 1

Answer: (B)

$5\sqrt{3} \, \text{N}$

Answer 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:

$5 \sqrt{3} \, \text{N}.$