Problem 373 - JEE Physics

Question

On the horizontal surface of a truck \((=0.6)\), a block of mass \(1 \mathrm{~kg}\) is placed. If the truck is accelerating at the rate of \(5 \mathrm{~m} / \mathrm{s}^{2}\) then frictional force on the block will be \(\mathrm{N}\) (A) 5 (B) 6 (C) \(5.88\) (D) 8

Step-by-Step Solution

Answer

The frictional force on the block is \(5.88\mathrm{N}\).

Step 1: Find the normal force on the block

To find the normal force on the block, we first need to find the weight of the block, which is the gravitational force acting on it: \[W = mg\] where \(W\) = weight of the block, \(m\) = mass of the block = \(1\mathrm{~kg}\), and \(g\) = acceleration due to gravity = \(9.8\mathrm{~m} / \mathrm{s}^{2}\). Substituting the values and calculating the weight: \[W = (1\mathrm{~kg})(9.8\mathrm{~m} / \mathrm{s}^{2}) = 9.8\mathrm{N}\] Since the block is resting on a horizontal surface and there is no vertical acceleration, the normal force (\(N\)) is equal to the weight of the block: \[N = W = 9.8\mathrm{N}\]

Step 2: Calculate the frictional force

To find the frictional force (\(f\)), we use the frictional force equation: \[f = μN\] where \(f\) = frictional force, \(μ\) = frictional coefficient = \(0.6\), and \(N\) = normal force = \(9.8\mathrm{N}\). Substituting the values and calculating the frictional force: \[f = (0.6)(9.8\mathrm{N}) = 5.88\mathrm{N}\] Hence, the frictional force on the block is \(5.88\mathrm{N}\). The correct answer is (C).

Key Concepts

Normal Force

The normal force is a key concept in understanding friction and motion. When an object like a block is resting on a surface, it experiences a force due to gravity called weight. This pulls it down towards the Earth's center. The normal force acts perpendicular to the surface, counterbalancing the weight of the object.
This means if there's no vertical motion, the normal force is usually equal to the weight. In equations, you often see it represented as:

• If the block has a mass (\(m\)) of\(1 \mathrm{~kg}\) and gravity (\(g\)) acts at\(9.8 \mathrm{~m/s}^{2}\), the weight (\(W\)) becomes:\[W = mg = 1 \times 9.8 = 9.8 \mathrm{~N}\].
• Since the surface is horizontal and there's no tilt or extra vertical movement, the normal force (\(N\)) equals the weight.

By understanding this balance, you can then explore how it interacts with other forces like friction, helping you predict how objects behave in different scenarios.

Coefficient of Friction

The coefficient of friction is a dimensionless number representing the interaction between two surfaces. It helps determine how easily one object slides over another. Higher values mean more resistance, while lower values imply smoother surfaces.
The coefficient of friction (\(μ\)) simplifies to an equation to find the frictional force:

• Frictional force (\(f\)) can be calculated as:\[f = μN\],where (\(N\)) is the normal force.
• For our block on the truck example, with\(μ = 0.6\), and\(N = 9.8 \, \mathrm{N}\):\[f = 0.6 \times 9.8 = 5.88 \, \mathrm{N}\].

The coefficient of friction varies based on the materials in contact, such as rubber on concrete being higher than wood on metal. Understanding these values is crucial for designing systems where precise motion control is needed or where minimizing resistance is desired.

Acceleration

Acceleration refers to the change in velocity of an object over time. It's a vector quantity, meaning it has both magnitude and direction. In the problem involving the truck, acceleration affects how forces like friction respond to the motion.
The truck accelerates at\(5 \mathrm{~m/s}^{2}\), impacting the block in two main ways:

• Dynamic forces on the block: As the truck speeds up, the friction must resist this acceleration to keep the block from slipping.
• Balancing forces: If the block was free of any friction, it would fall off the truck. But because of the frictional force calculated, the block stays put.

Understanding acceleration allows students to predict how and why certain changes in motion occur under different force conditions. The interaction of acceleration with friction in this scenario shows how crucial balance between forces is to maintaining steady motion of objects.