Question

On the horizontal surface of a truck a block of mass 1 kg is placed (µ = 0.6) and truck is moving with acceleration 5 m/s² then the frictional force on block will be : -

• A. 5N
• B. 6N
• C. 5.88N
• D. 8N

Answer 1

The Correct Option is A

To determine the frictional force acting on the block placed on the horizontally accelerating truck, we need to understand the dynamics involved. Here, we're given:

• Mass of the block, m = 1 \, \text{kg}
• Coefficient of friction, \mu = 0.6
• Acceleration of the truck, a = 5 \, \text{m/s}^2

The frictional force can be calculated using the formula:

f = \mu \cdot N

Where N is the normal force. On a horizontal surface, the normal force, N, is equal to the gravitational force acting on the block:

N = m \cdot g

Substituting the values:

N = 1 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N}

Now, let's calculate the maximum static frictional force:

f_{\text{max}} = \mu \cdot N = 0.6 \times 9.8 = 5.88 \, \text{N}

The truck is accelerating, causing a horizontal force on the block. This force can be expressed using Newton's second law:

F_{\text{horizontal}} = m \cdot a = 1 \, \text{kg} \times 5 \, \text{m/s}^2 = 5 \, \text{N}

Since the block can only resist up to 5.88 \, \text{N} of frictional force from moving, and the horizontal force due to the acceleration of the truck is 5 \, \text{N}, the actual frictional force acting on the block will be exactly 5 \, \text{N}. Therefore, the block does not slip.

The correct answer is 5N, as the frictional force supports the motion without exceeding the maximum static friction.