91Ó°ÊÓ

When objects are at rest on a surface and there's a resistance that prevents them from moving, we're dealing with static friction. Static friction is a force that holds objects in place and stops them from sliding. In the context of the flatcar with crates, static friction is what's stopping the crates from sliding while the train decelerates.

To calculate static friction, we use the following formula:

• The maximum static friction force, \( f_s \), is given by: \( f_s = \mu_s \cdot m \cdot g \).

Here, \( \mu_s \) is the coefficient of static friction, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \mathrm{m/s^2} \)). This equation helps us determine the force needed to overcome static friction and initiate motion.

The coefficient of static friction, \( \mu_s \), varies between surfaces and is a measure of how much frictional force they can exert. For our exercise, \( \mu_s = 0.25 \). This means the surface is somewhat slippery, but not entirely frictionless.

Acceleration is the rate at which an object's velocity changes over time. When the train decelerates to stop without the crates sliding, the acceleration must be less than or equal to the maximum provided by static friction.

We know that the maximum acceleration is dictated by the equation:

• \( a = \mu_s \cdot g \)
• With \( \mu_s = 0.25 \) and \( g = 9.8 \, \mathrm{m/s^2} \), the maximum acceleration \( a \) is \( 2.45 \, \mathrm{m/s^2} \).

This means the train must decelerate at this rate or slower to ensure that the crates do not slide. In simple terms, the force exerted by static friction must be able to counteract the change in velocity from deceleration.

Understanding acceleration in this context helps us manage how forces interact when objects come to a stop, ensuring stability and safety.

Kinematic equations are crucial tools for analyzing motion in physics. They allow us to relate the basic parameters of motion: displacement, initial and final velocities, acceleration, and time. In this exercise, we're looking to find the stopping distance using these equations.

The relevant kinematic equation here is:

• \( v^2 = u^2 + 2a s \)
• where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the displacement or stopping distance.

Given that the train's final velocity \( v \) is \( 0 \, \mathrm{m/s} \) and initial velocity \( u \) is \( 13.33 \, \mathrm{m/s} \), we can rearrange the equation to solve for the stopping distance \( s \).

By plugging in the values and solving, we find the stopping distance to be \( 36.2 \, \mathrm{meters} \). This application of kinematic equations is pivotal to solving motion problems where time isn't directly involved.