Problem 99 - Physics for Scientists and Engineers with Modern Physics

Question

A \(72-\mathrm{kg}\) 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 \(\mu_{\mathrm{k}}=0.25\) (Fig. \(5-55)\). (a) What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude \(F_{\mathrm{T}}=240 \mathrm{~N}\) 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 \(F_{\mathrm{T}}=240 \mathrm{~N}\) 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) The skier's acceleration is approximately 0.883 m/s^2. (b) The skier's acceleration is approximately 0.987 m/s^2. (c) The angled force reduces friction, increasing net force and acceleration.

Step 1: Understand the Forces on the Skier

The skier is being affected by several forces: the tension in the rope, the gravitational force, and the kinetic friction. The horizontal force due to tension is either directly horizontal (part a) or at an angle (part b), while the vertical forces include the skier's weight and the vertical component of the tension.

Step 2: Calculate Net Force for Part (a)

In part (a), the tension is directly horizontal, so the components are: horizontal force from tension, \( F_{\text{T}} = 240\, \text{N} \) and frictional force, \( f_k = \mu_k \cdot F_{\text{N}} = \mu_k \cdot mg = 0.25 \times 72 \times 9.8 = 176.4\, \text{N} \). The net force is: \( F_{\text{net}} = F_{\text{T}} - f_k = 240 - 176.4 = 63.6\, \text{N} \).

Step 3: Calculate Acceleration for Part (a)

Using Newton’s second law, \( F = ma \), we can solve for acceleration: \( a = \frac{F_{\text{net}}}{m} = \frac{63.6}{72} \approx 0.883\, \text{m/s}^2 \).

Step 4: Determine Forces for Part (b)

Now, the tension force is applied at an angle, thus it has a horizontal component \( F_{\text{T}x} = 240 \cos(12^Degrees) \) and a vertical component. Calculate these: \( F_{\text{T}x} = 240 \cos(12^Degrees) \approx 234.6\, \text{N} \). The vertical component affects the normal force: \( F_{\text{N}} = mg - F_{\text{T}y} = 72 \times 9.8 - 240 \sin(12^Degrees) \approx 703.9 - 49.7 = 654.2\, \text{N} \).

Step 5: Calculate Net Force and Acceleration for Part (b)

Calculate the frictional force using the adjusted normal force: \( f_k = \mu_k \cdot F_{\text{N}} = 0.25 \times 654.2 = 163.55\, \text{N} \). The net force is now \( F_{\text{net}} = F_{\text{T}x} - f_k \approx 234.6 - 163.55 = 71.05\, \text{N} \). Solving for acceleration: \( a = \frac{71.05}{72} \approx 0.987\, \text{m/s}^2 \).

Step 6: Interpret Results for Part (c)

In part (b), the tension applied at an angle decreases the effective gravitational force acting against the skier's motion, thus reducing the kinetic friction. This reduction in opposing force results in a greater net force, and consequently, a larger acceleration.

Key Concepts

Kinetic Friction

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

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 Calculation

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.