91Ó°ÊÓ

Kinetic friction is a force that opposes the motion of two surfaces sliding past each other. It's what stops objects from sliding easily and keeps them in place. In the context of our problem, when the box slides down the ramp, kinetic friction slows it down.

The formula for kinetic friction is given by:

• \( f_k = \mu_k \times N \).
• Here, \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force exerted by the ramp on the box.

In our exercise, this coefficient is 0.300, which tells us how strongly the surfaces resist movement. The normal force is calculated considering the weight of the object and the angle of the incline, as well as any other forces acting on the box, such as the horizontal force in this case. Substituting the numbers gives us the frictional force, which is \( 16.931 \, \text{N} \), showing that friction plays a significant role in the problem.

Newton's Second Law is crucial in understanding how forces affect motion. It states that the acceleration of an object depends on the net force acting on it and its mass:

• \( F = ma \).
• In this equation, \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.

When analyzing the motion of the box on the ramp, all forces acting on the box, such as gravity, friction, and the applied horizontal force, contribute to the net force. The forces parallel to the incline are combined to find the total force that pushes or pulls the box down the ramp. Once we have the net force, it's divided by the box's mass to find the acceleration. In this situation, the calculated acceleration is \( 2.76 \, \text{m/s}^2 \), showing how quickly the box speeds up as it moves down the ramp.

Kinematics is the branch of physics that describes the motion of objects. It uses equations to quantify how fast, far, or long an object moves. In our problem, kinematics helps us find the final speed of the box after a given time using the formula:

• \( v = u + at \).
• Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

Since the box starts from rest, our initial velocity \( u \) is 0. The acceleration \( a \) was computed earlier as \( 2.76 \, \text{m/s}^2 \). By plugging these values into the equation along with time \( t = 2 \, \text{s} \), we get a final speed of \( 5.52 \, \text{m/s} \). This calculation clarifies how quickly the box moves down the ramp after 2 seconds.