91Ó°ÊÓ

When calculating the frictional force acting on an object, it's essential to understand that frictional force resists the motion of the object sliding over a surface. In the exercise provided, the frictional force, \( \mathbf{F}_{f} \), is determined using the coefficient of kinetic friction \( \mu_{k} \) and the normal force \( \mathbf{F}_{n} \).
The formula is:

• \( \mathbf{F}_{f} = \mu_{k} \cdot \mathbf{F}_{n} \)

Since the crate rests on a flat surface and no external vertical forces are applied, \( \mathbf{F}_{n} \) is equal to the weight of the crate, \( \mathbf{W} \).
The weight, in turn, is the product of the mass of the crate and gravitational acceleration:

• \( \mathbf{W} = m \cdot g \)

Putting in the given values:

• \( \mathbf{F}_{f} = 0.3 \cdot (50 \text{ kg} \cdot 9.8 \text{ m/s}^2) = 147 \text{ N} \)

This value, 147 N, represents the force opposing the motion due to friction. It plays a crucial part in understanding how much additional force \( \mathbf{P} \) is required to move the crate at the desired acceleration.

Newton's Second Law of Motion is a fundamental principle used to relate the forces acting on an object to its acceleration. This law is expressed as:
\[ \mathbf{F} = \mathbf{m} \cdot \mathbf{a} \]
where \( \mathbf{F} \) is the net force acting on the object, \( \mathbf{m} \) is the mass, and \( \mathbf{a} \) is the acceleration. In the dynamics problem involving the crate, we need to find the force \( \mathbf{P} \) acting on it.

Let's break down the forces:

• The crate has a resultant force \( \mathbf{P} - \mathbf{F}_{f} \) where \( \mathbf{P} \) is the applied force and \( \mathbf{F}_{f} \) is the frictional force, as friction acts in the opposite direction of \( \mathbf{P} \).
• This resultant force equals the product of mass and acceleration from Newton's second law.

Reorganizing gives:

• \( \mathbf{P} = \mathbf{F}_{f} + (\mathbf{m} \cdot \mathbf{a}) \)

Using the mass \( 50 \text{ kg} \) and calculated acceleration \( 0.8 \text{ m/s}^2 \), we find \( \mathbf{P} \):

• \( \mathbf{P} = 147 \text{ N} + (50 \text{ kg} \cdot 0.8 \text{ m/s}^2) = 187 \text{ N} \)

Therefore, the total force required to move the crate, considering friction, is 187 N.

Kinematic equations connect the various parameters of motion, such as velocity, acceleration, time, and distance, without considering the causes of motion (forces). They are essential tools in physics for solving problems related to motion.

For the scenario of the crate, we employ the equation:

• \( v^2 = u^2 + 2as \)

This equation relates initial velocity \( u \), final velocity \( v \), acceleration \( a \), and distance \( s \) traveled.

Here are the given values:

• Initial velocity \( u = 0 \) (since the crate starts from rest),
• Final velocity \( v = 4 \text{ m/s} \),
• Distance \( s = 5 \text{ m} \).

Rearrange the equation to solve for acceleration \( a \):

• \( a = \frac{(v^2 - u^2)}{2s} \)

Substitute the known values:

• \( a = \frac{(4 \text{ m/s})^2 - (0 \text{ m/s})^2}{2 \cdot 5 \text{ m}} = 0.8 \text{ m/s}^2 \)

This result, \( 0.8 \text{ m/s}^2 \), is the acceleration of the crate, crucial for determining the net force required (as seen in Newton's second law).