91Ó°ÊÓ

The force of friction is a force that opposes the motion of two surfaces sliding against each other. In our snowmobile scenario, the snowmobile moves at a constant velocity, indicating a balance between the drive force and the force of friction.
_Key characteristics of friction:_

• Friction acts in the opposite direction of motion.
• When the drive force is equal to the force of friction, the velocity is constant.

When the snowmobile's drive force of \(205 \, \mathrm{N}\) was applied, the frictional force was also \(205 \, \mathrm{N}\). This balance shows that there is no net force acting on the snowmobile, allowing it to move steadily without accelerating or decelerating. Once the drive force is removed, friction becomes the sole force, causing the snowmobile to slow down until it stops.

In kinematics, Newton's second law is crucial to determine how objects move. The law states: _The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass_. The formula is given as \( f = ma \), where \(f\) is the force applied, \(m\) is the mass, and \(a\) is the acceleration.
For the snowmobile, the force of friction now is the unbalanced force once the drive force stops. Using Newton's second law, you can find the magnitude of deceleration that this frictional force causes:

• Given: \( \text{Force of friction} = 205 \, \mathrm{N} \), \( \text{Mass} = 136 \, \mathrm{kg} \)
• The deceleration is \( a = \frac{-205}{136} \approx -1.51 \, \mathrm{m/s^2} \)

The negative sign indicates that this is a slowing effect (deceleration), not an acceleration.

Kinematic equations are essential for solving problems involving motion, where constant acceleration or deceleration is present. These equations relate initial velocity, final velocity, time, distance, and acceleration.
The two primary kinematic equations used in our problem are:

• \( v^2 = u^2 + 2as \) (to find distance)
• \( v = u + at \) (to find time)

Applying these:
1. **Distance calculation:** When the initial speed \(u\) is \(5.50 \, \mathrm{m/s}\), final velocity \(v\) is \(0 \, \mathrm{m/s}\), and deceleration \(a = -1.51 \, \mathrm{m/s^2}\), solve for \(s\):
\[ s = \frac{(5.50)^2}{2 \times 1.51} \approx 9.98 \, \mathrm{m} \]
2. **Time to stop:** When \(a = -1.51 \, \mathrm{m/s^2}\), \(u = 5.50 \, \mathrm{m/s}\), and \(v = 0\), solve for \(t\):
\[ t = \frac{0 - 5.50}{-1.51} \approx 3.64 \, \mathrm{s} \]
This approach gives you the distance over which the snowmobile comes to a stop and the time duration of deceleration.