To solve this problem, we analyze the forces acting on the system composed of the block and the wedge.

1. Acceleration of the Block

First, we determine the acceleration a of the block m as it slides down the wedge. The block starts from rest ($v_0=0$) and reaches a velocity v after traveling a distance s. We can use the kinematic equation:

$$v^2 = v_0^2 + 2as$$

Solving for the acceleration a:

$$a = \frac{v^2}{2s}$$

Substituting the given values:

$s = 1.4$ m, $v = 1.4$ m/s. $$a = \frac{(1.4 \text{ m/s})^2}{2(1.4 \text{ m})} = \frac{1.4}{2} \text{ m/s}^2 = 0.7 \text{ m/s}^2$$

2. System Dynamics and Friction Force

[Q1] To find the friction force exerted by the ground on the wedge, we consider the block and the wedge as a single system. The wedge is stationary, so its acceleration is zero. The block has an acceleration a directed down the inclined plane.

Let's apply Newton's second law for a system of particles in the horizontal direction:

$$\sum F_{\text{ext}, x} = m a_{m,x} + M a_{M,x}$$

where:

• $\sum F_{\text{ext}, x}$ is the sum of external horizontal forces on the system. The only external horizontal force is the static friction from the ground on the wedge, $f_s$.
• $a_{m,x}$ is the horizontal component of the block's acceleration.
• $a_{M,x}$ is the horizontal component of the wedge's acceleration, which is zero since the wedge is stationary.

From the figure, the block slides down and to the left. Let's define the positive x-axis to the right. The block's acceleration vector a points down the incline at an angle \theta to the horizontal. Its horizontal component is:

$$a_{m,x} = -a \cos\theta$$

The negative sign indicates the acceleration is to the left.

Substituting into Newton's second law:

$$f_{s,x} = m(-a \cos\theta) + M(0)$$ $$f_{s,x} = -ma \cos\theta$$

The magnitude of the friction force is therefore:

$$f_s = |f_{s,x}| = ma \cos\theta$$

Now, we substitute the numerical values:

$m = 1.0$ kg, $a = 0.7$ m/s², $\theta = 30^\circ$. $$f_s = (1.0 \text{ kg})(0.7 \text{ m/s}^2) \cos(30^\circ) = 0.7 \times \frac{\sqrt{3}}{2} \text{ N}$$ $$f_s = 0.35\sqrt{3} \text{ N} \approx 0.61 \text{ N}$$

[Q2] The direction of the friction force is given by the sign of $f_{s,x}$. Since $m, a, \cos\theta$ are all positive, $f_{s,x}$ is negative. This means the friction force exerted by the ground on the wedge points in the negative x-direction, which is to the left.

This result is counter-intuitive. One might expect the block sliding left to push the wedge left, requiring a friction force to the right. However, a detailed analysis shows the block's acceleration causes the wedge to be pushed to the right, necessitating a friction force to the left. The force the block exerts on the wedge has a horizontal component ma cos(theta) to the right. The static friction must balance this, so it acts to the left with the same magnitude.