To solve this problem, we need to determine the velocity of a crate moving up an inclined plane after it travels 6 meters, starting from rest. The problem provides the following information:

• Gravitational force $g = 9.81 \, \text{m/s}^2$
• Mass of the crate $m = 50 \, \text{kg}$
• Coefficient of kinetic friction $\mu_k = 0.25$
• Displacement along the inclined plane $d = 6 \, \text{m}$
• The crate starts from rest, so the initial velocity is $v_0 = 0 \, \text{m/s}$

Step-by-step approach:

1. Forces on the crate:

   • Gravitational force component along the incline:
     $F_g = m \cdot g \cdot \sin(\theta)$, where $\theta$ is the angle of the incline.

   • Frictional force:
     $F_f = \mu_k \cdot N$, where $N$ is the normal force. On an incline, $N = m \cdot g \cdot \cos(\theta)$.

2. Net force: The net force $F_{\text{net}}$ acting on the crate is the sum of the gravitational force and the frictional force, both acting opposite to the direction of motion (up the incline): $F_{\text{net}} = - F_g - F_f$ or $F_{\text{net}} = - m \cdot g \cdot \sin(\theta) - \mu_k \cdot m \cdot g \cdot \cos(\theta)$

3. Acceleration using Newton’s second law: From Newton’s second law, $F = m \cdot a$, so the acceleration $a$ can be found by dividing the net force by the mass of the crate: $a = \frac{F_{\text{net}}}{m} = - g \cdot (\sin(\theta) + \mu_k \cdot \cos(\theta))$

4. Kinematic equation to find the velocity: Use the following kinematic equation, which is appropriate since the crate starts from rest: $v^2 = v_0^2 + 2 \cdot a \cdot d$ Since $v_0 = 0$, the equation simplifies to: $v = \sqrt{2 \cdot a \cdot d}$

Assumptions:

We would need to know the angle $\theta$ of the incline to proceed with the calculations. If the angle is not given, we might need to make an assumption or it could be provided in Figure 1.

Would you be able to provide the angle of the incline or any additional information from the figure?