Problem 154 - A Complete Resource Book in Physics for JEE Main

Question

A block rests on a rough inclined plane making an angle of \(30^{\circ}\) with the horizontal. The co-efficient of static friction between the block and the plane is \(0.8\). If the frictional force on the block is \(10 \mathrm{~N}\), the mass of the block (in \(\mathrm{kg}\) ) is (Taking \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(2.0\) (B) \(4.0\) (C) \(1.6\) (D) \(2.5\)

Step-by-Step Solution

Answer

The mass of the block is approximately \(1.6 \mathrm{~kg}\) (C).

Step 1: Draw a force diagram

Draw a force diagram for the block, which includes the weight of the block (mg), normal force (N) perpendicular to the inclined plane, and frictional force (f) opposing the block's motion.

Step 2: Calculate the frictional force

The frictional force (f) can be calculated using the formula: \(f = \mu N\) where \(\mu\) is the coefficient of static friction and \(N\) is the normal force.

Step 3: Calculate the normal force

To find the normal force (N), we need to break down the weight of the block into two components: component parallel to the incline (\(mg\sin{\theta}\)), and component perpendicular to the incline (\(mg\cos{\theta}\)). The normal force is equal to the component of the weight perpendicular to the incline: \(N = mg\cos{\theta}\)

Step 4: Substitute the values

Now substitute the given values in the equation: \(f = \mu N\) \(10 = 0.8 \times (m)(10\cos{30^{\circ}})\)

Step 5: Solve for the mass (m)

Now we can solve for the mass of the block (m): \(10 = 0.8 \times (m)(10\times(\frac{\sqrt{3}}{2}))\)

Step 6: Simplify and select the correct answer

Simplify the equation and solve for m: \(10 = 4\sqrt{3} \times m\) \(m = \frac{10}{4\sqrt{3}}\) \(m = \frac{5}{2\sqrt{3}}\) \(m = \frac{5\sqrt{3}}{6}\) The closest answer from the options is \(1.6 \mathrm{~kg}\) (C).