Speed and Distance Traveled in an Interesting Exercise

What is the speed of the small block after reaching equilibrium?

How far does the slab travel before reaching the speed of the small block?

Answer:

The speed of the small block after reaching equilibrium is v₁ = 0.375 m/s.

The distance traveled by the slab before reaching the speed of the small block is x = 0.335 m.

Let's analyze this interesting exercise. The block moves with a constant speed and experiences friction force with the slab. The friction force opposes the movement of the block and is exerted in the direction of the slab's movement.

In this scenario, the system formed by the block and the slab reaches a stable speed where friction cannot provide any more energy to the system. This equilibrium speed can be determined using the conservation of linear momentum.

By applying Newton's second law to the system, we can find the speed of the small block after reaching equilibrium:

v₁ = m / (m + M) * v₀

Initially, the slab is at rest so its initial velocity is zero. Using kinematics, we can calculate the distance traveled by the slab before reaching the speed of the small block:

x = v₁² / (2 * a)

By applying Newton's second law to the block and the slab, we can determine the friction force and acceleration. The friction force between the block and the slab is μ * m * g, and the acceleration of the slab is μ * g * m / M.

Calculating the acceleration and distance traveled, we find that the speed of the small block after reaching equilibrium is 0.375 m/s, and the distance traveled by the slab is 0.335 m.