91Ó°ÊÓ

When an object moves across a surface, it encounters a force called kinetic friction. This force acts opposite to the direction of movement. In the case of our water skier, the skis glide over the water, and kinetic friction works against the forward pull from the boat. It’s crucial in determining how easily the skier can be accelerated.

Kinetic friction depends on two primary factors:

• The coefficient of kinetic friction (\( \mu_k \)), which is a measure of how slippery or sticky the two surfaces are relative to each other. For our skier, this value is 0.25.
• The normal force (\( F_N \)), which is the perpendicular force exerted by the surface to support the weight of the object. Here, it's the gravitational force experienced by the skier.

The formula to calculate the kinetic frictional force is:\[ f_k = \mu_k \cdot F_N \]Applying this to part (a) of our problem, with the gravitational force being equal to the skier's weight (\( mg \)), the skier experiences a frictional force of\( 176.4 \, \text{N} \)as it slides over the water.

Tension force emerges from a rope, cable, or similar object when it is pulled tight by forces acting from opposite ends. In this exercise, the tension in the rope is what pulls the skier forward.

For part (a) of our problem, the tension force is horizontal, matching the angle of the boat's pull:

• Magnitude of tension, \( F_T = 240 \, \text{N} \)
• Direction, \( \theta = 0^\circ \) , meaning the force is directly in line without vertical components.

In part (b), the tension force is applied at a 12° angle. This introduces both a horizontal component (\( F_{Tx} \)) and a vertical component (\( F_{Ty} \)):

• Horizontal component: \( F_{Tx} = 240 \cos(12^\circ) \approx 234.6 \, \text{N} \)
• Vertical component: \( F_{Ty} = 240 \sin(12^\circ) \approx 49.7 \, \text{N} \)

The vertical tension force slightly lifts the skier, reducing the normal force and thus the kinetic friction. This results in clearer motion and contributes to a greater net acceleration.

Acceleration is a crucial aspect of motion governed by Newton's second law of motion. It describes how quickly an object changes its velocity in response to net forces acting on it.

In our scenario, the net force (\( F_{net} \)) is the result of subtracting the frictional force from the horizontal component of the tension force. For part (a), the calculation is straightforward since the force is entirely horizontal:\[ F_{net} = F_T - f_k = 240 \, \text{N} - 176.4 \, \text{N} = 63.6 \, \text{N} \]Using Newton's formula, \( F = ma \), the acceleration is derived:\[ a = \frac{F_{net}}{m} = \frac{63.6}{72} \approx 0.883 \, \text{m/s}^2 \]
For part (b), the introduction of an angle changes the calculations slightly. Adjusting for both components of the tension force results in:\[ F_{net} = F_{Tx} - f_k \approx 234.6 \, \text{N} - 163.55 \, \text{N} = 71.05 \, \text{N} \]The skier's acceleration becomes:\[ a = \frac{71.05}{72} \approx 0.987 \, \text{m/s}^2 \]
It is apparent that by reducing the opposing kinetic force through the added incline effect, the skier accelerates faster with less friction to hinder the movement.