Problem 99 - Physics for Scientists and Engineers with Modern Physics

Question

A 72 -kg water skier is being accelerated by a ski boat on a flat \((\) "glassy") lake. The coefficient of kinetic friction between the skier's skis and the water surface is \(\mu_{k}=0.25\) (Fig. 55 ). (a) What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude \(F_{T}=240 \mathrm{N}\) to the skier \(\left(\theta=0^{\circ}\right) ?\) (b) What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of \(F_{T}=240 \mathrm{N}\) on the skier at an upward angle \(\theta=12^{\circ} ?\) (c) Explain why the skier's acceleration in part \((b)\) is greater than that in part \((a)\) .

Step-by-Step Solution

Answer

(a) 0.883 m/s²; (b) 0.982 m/s²; (c) Less friction with angled tension increases acceleration.

Step 1: Analyze forces acting on the skier (Part a)

For part (a), the forces acting on the skier are the horizontal tension force pulling him and the kinetic friction opposing his motion. The tension force is given as \( F_T = 240 \, \text{N} \) and acts purely horizontally. The normal force (\( N \)) is equal to the gravitational force, since there is no vertical component of the tension. Thus, \( N = mg = 72 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 705.6 \, \text{N} \). The frictional force (\( f_k \)) is \( \mu_k N = 0.25 \times 705.6 \, \text{N} = 176.4 \, \text{N} \).

Step 2: Calculate net force and acceleration (Part a)

The net force acting on the skier is the tension force minus the frictional force. \( F_{\text{net}} = F_T - f_k = 240 \, \text{N} - 176.4 \, \text{N} = 63.6 \, \text{N} \). From Newton's second law, the acceleration \( a \) is given by \( F_{\text{net}} = ma \). Therefore, \( a = \frac{F_{\text{net}}}{m} = \frac{63.6 \, \text{N}}{72 \, \text{kg}} = 0.883 \, \text{m/s}^2 \).

Step 3: Analyze forces with angle (Part b)

In part (b), since the rope is pulling at an angle \( \theta = 12^\circ \), the tension has both horizontal and vertical components. The horizontal component of the tension is \( F_{T_x} = F_T \cos(\theta) = 240 \, \text{N} \times \cos(12^\circ) \approx 234.52 \, \text{N} \). The vertical component reduces the normal force: \( N = mg - F_T \sin(\theta) = 705.6 \, \text{N} - (240 \, \text{N} \times \sin(12^\circ)) \approx 655.1 \, \text{N} \). The frictional force is \( f_k = \mu_k N = 0.25 \times 655.1 \, \text{N} \approx 163.78 \, \text{N} \).

Step 4: Calculate net force and acceleration (Part b)

With the horizontal tension component and adjusted friction, the net force is \( F_{\text{net}} = F_{T_x} - f_k = 234.52 \, \text{N} - 163.78 \, \text{N} = 70.74 \, \text{N} \). Using Newton's second law, \( a = \frac{F_{\text{net}}}{m} = \frac{70.74 \, \text{N}}{72 \, \text{kg}} \approx 0.982 \, \text{m/s}^2 \).

Step 5: Explain why acceleration increases in Part b

The skier's acceleration in part (b) is greater than in part (a) because the upward angle of the tension force reduces the normal force, thus reducing the kinetic friction opposing the skier's motion. With less friction to overcome, the net force is larger, resulting in greater acceleration.

Key Concepts

Kinetic Friction

Kinetic friction occurs when two surfaces slide against each other and resist motion. In the context of a water skier, the skis sliding on water experience kinetic friction. The force exerted by kinetic friction can be calculated using the formula:

• \( f_k = \mu_k \cdot N \)

where \( f_k \) is the kinetic frictional force, \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force, which is the force perpendicular to the surfaces in contact.
For our water skier, the essential point is how kinetic friction works against the skier's motion. This friction depends on both the weight of the skier (as part of the normal force) and the characteristics of the ski and water surfaces (represented by \( \mu_k \)). The goal in analyzing friction in scenarios like this is to understand how it reduces the skier's acceleration by opposing the tension force applied by the boat.

Tension Force

The tension force is the pulling force transmitted through a string, rope, or cable when it is pulled tight by forces acting from opposite ends. For the water skier, this force is applied through the rope that connects him to the boat.
The tension in the rope helps overcome friction and accelerates the skier. When the rope pulls directly in line with the skier (horizontally), it provides maximum forward force.

• In part (b) of the problem, the rope exerts force at an angle, splitting into horizontal and vertical components:
• Horizontal tension, \( F_{T_x} = F_T \cos(\theta) \)
• Vertical tension, \( F_{T_y} = F_T \sin(\theta) \)

These components affect the acceleration because they also impact the normal force and consequently the friction. Different angles lead to differing distributions of force, influencing the skier's speed and trajectory.

Net Force Calculation

Net force is the overall force acting on an object when all individual forces are combined. For the skier, determining net force involves balancing the tension force with the opposing force of kinetic friction. Using Newton's second law, we relate net force to an object's mass and acceleration:

• \( F_{\text{net}} = ma \)

In our exercise:- The skier's acceleration is found by taking the difference between the tension and frictional forces.- For part (a), the direct application of Newton's second law gives us \( F_{\text{net}} = F_T - f_k \).- In part (b), calculating net force also involves accounting for the tension angle, which changes the effective horizontal force of tension.
By understanding these calculations, you can see how different forces contribute to the skier's motion.

Newton's Laws of Motion

Newton's Laws of Motion provide the foundation for analyzing motion. For the moving skier, the first law of motion explains that an object at rest stays at rest unless acted upon by a force. The skier initially resists motion due to friction but accelerates when pulled by the rope.
The second law is most essential here, which states that the acceleration of an object depends on the net force acting on it and its mass. The formula is:

• \( F_{\text{net}} = ma \)

This principle is used in our calculations to figure out the skier's acceleration from net force.

• In part (a) and part (b) of the exercise, we use this law to assess how forces like tension and friction shape the skier's movement.

The third law isn’t directly solved but still important, reminding us that forces come in pairs—meaning the skier pulls back on the rope just as hard as the boat pulls on him.
Comprehending these laws helps illuminate the skiier's physical experience and guides problem-solving in physics challenges.