The block of mass 2kg and 3 kg are placed over each other. The friction between blocks is 0.2 and the friction between 3 kg block and horizontal surface is 0.06. A force of F=t/2 (t is in seconds) is applied on upper block in rightward direction.1. The relative slipping between blocks occurs at t=? 2. Friction force acting between two blocks at t=8seconds

To solve the problem involving the two blocks and the forces acting on them, we need to analyze the forces and the conditions for slipping. Let's break this down step by step.

Understanding the Forces at Play

We have two blocks: a 2 kg block on top of a 3 kg block. The coefficients of friction are given as follows:

• Friction between the blocks (μ₁): 0.2
• Friction between the 3 kg block and the surface (μ₂): 0.06

A force F is applied to the upper block, which varies with time as F = t/2, where t is in seconds. Our goal is to determine:

1. The time at which relative slipping occurs between the blocks.
2. The friction force acting between the two blocks at t = 8 seconds.

Calculating the Maximum Static Friction

First, we need to find the maximum static friction that can act between the two blocks. The normal force (N₁) acting on the 2 kg block is equal to its weight:

N₁ = m₁ * g = 2 kg * 9.81 m/s² = 19.62 N

The maximum static friction (f_max) between the blocks can be calculated using:

f_max = μ₁ * N₁ = 0.2 * 19.62 N = 3.924 N

Finding the Force Applied on the Upper Block

The force applied on the upper block increases with time. At any time t, the force is:

F(t) = t/2

To find when slipping occurs, we need to determine when the applied force exceeds the maximum static friction. This happens when:

F(t) > f_max

Substituting the expressions, we have:

t/2 > 3.924 N

Multiplying both sides by 2 gives:

t > 7.848 seconds

Thus, relative slipping between the blocks occurs at:

t ≈ 7.85 seconds

Calculating the Friction Force at t = 8 Seconds

Now, let's find the friction force acting between the blocks at t = 8 seconds. At this moment, the applied force is:

F(8) = 8/2 = 4 N

Since 4 N is less than the maximum static friction of 3.924 N, the blocks do not slip at this point. Therefore, the friction force will be equal to the applied force:

Friction force at t = 8 seconds = 4 N

Summary of Results

In summary, we found that:

• Relative slipping between the blocks occurs at approximately t = 7.85 seconds.
• The friction force acting between the two blocks at t = 8 seconds is 4 N.

This analysis illustrates the importance of understanding the forces involved and how they interact over time. If you have any further questions or need clarification on any part of this problem, feel free to ask!