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Chapter 7: Problem 3

\(\cdot\) A boat with a horizontal tow rope pulls a water skier. She skis off to the side, so the rope makes an angle of \(15.0^{\circ}\) with the forward direction of motion. If the tension in the rope is \(180 \mathrm{N},\) how much work does the rope do on the skier during a forward displacement of 300.0 \(\mathrm{m} ?\)

Short Answer

Expert verified
The work done is approximately 52158.6 J.

Step by step solution

01

Define Work Formula

The work done by a force is given by the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( W \) is the work, \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and the direction of displacement.
02

Identify Known Values

From the problem, the tension in the rope \( F = 180 \, \text{N} \), the forward displacement \( d = 300.0 \, \text{m} \), and the angle \( \theta = 15.0^{\circ} \).
03

Calculate Work Using Given Values

Substitute the known values into the work formula: \[ W = 180 \, \text{N} \times 300.0 \, \text{m} \times \cos(15.0^{\circ}) \].
04

Compute Cosine of the Angle

Calculate \( \cos(15.0^{\circ}) \), which is approximately \( 0.9659 \).
05

Final Calculation of Work

Multiply the values together: \[ W = 180 \, \text{N} \times 300.0 \, \text{m} \times 0.9659 \]. This results in \( W \approx 52158.6 \, \text{J} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Formula
Understanding the work formula helps us determine how much energy is transferred by a force acting over a distance. The work done by a force is given by the formula:\[ W = F \cdot d \cdot \cos(\theta) \]Here,
  • \( W \) represents the work done in Joules (J).
  • \( F \) is the magnitude of the force applied in Newtons (N).
  • \( d \) is the displacement, or the distance moved by the object, measured in meters (m).
  • \( \theta \) is the angle between the force vector and the direction of displacement, measured in degrees.
By decomposing the vector into its components, the work formula accounts not only for the magnitude of the force and distance but also how aligned the force is with the direction of displacement. This insight is key to understanding the effectiveness of the force in doing work.
Force and Displacement
To better grasp how force relates to displacement, consider force as a push or pull applied to an object. Displacement, on the other hand, is how far the object moves and in what direction. In our example with the water skier, the rope applies a force of 180 N which pulls the skier 300.0 m in the forward direction.
However, whenever you consider work being done, it鈥檚 crucial to also think about the direction in which the force is applied. If the force and displacement are in the same direction, the full force contributes to doing the work. If the force is only partially aligned with the displacement, such as forming an angle with the path, only a portion of the force effectively contributes to moving the object forward. This nuance is captured in our work formula by the cosine of the angle between force and displacement.
Angle of Force Application
The angle of force application is a critical factor in determining the actual work done by a force. In the context of work, \( \theta \) is the angle between the force vector and the direction of the displacement. The angle affects how much of the force actually contributes to the movement of the object.
A

Role of Cosine in the Work Formula

Cosine is used in the work formula to account for how much of the force is effectively applied in the direction of the displacement. When the angle is zero (meaning the force is perfectly aligned with the displacement), \( \cos(\theta) \) is 1, so all the force contributes to work done. As the angle increases, \( \cos(\theta) \) decreases, and thus, less of the force contributes to the actual work.
If we return to our example, the tow rope makes a \( 15.0^{\circ} \) angle with the direction of motion. Therefore, the calculation involves multiplying the force by \( \cos(15.0^{\circ}) \), which is approximately 0.9659. This means that around 96.59% of the force is effective in doing work forward. Understanding this relationship between angle and work helps us calculate the energy efficacy in practical scenarios.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be \(10 \mathrm{g},\) and assume the wings move an average downward distance of 1.0 \(\mathrm{cm}\) during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

\(\bullet\) Volcanoes on Io. Io, a satellite of Jupiter, is the most volcanically active moon or planet in the solar system. It has volcanoes that send plumes of matter over 500 \(\mathrm{km}\) high (see the accompanying fig- ure). Due to the satellite's small mass, the acceleration due to gravity on Io is only \(1.81 \mathrm{m} / \mathrm{s}^{2},\) and \(\mathrm{Io}\) has no appreciable atmosphere. As- sume that there is no varia- tion in gravity over the distance traveled. (a) What must be the speed of material just as it leaves the volcano to reach an altitude of 500 \(\mathrm{km} ?\) (b) If the gravitational potential energy is zero at the surface, what is the potential energy for a 25 kg fragment at its maximum height on Io? How much would this gravitational potential energy be if it were at the same height above earth?

\(\bullet\) \(\bullet\) Automobile accident analysis. In an auto accident, a car hit a pedestrian and the driver then slammed on the brakes to stop the car. During the subsequent trial, the driver's lawyer claimed that the driver was obeying the posted 35 mph speed limit, but that the limit was too high to enable him to see and react to the pedestrian in time. You have been called as the state's expert witness. In your investigation of the accident site, you make the following measurements: The skid marks made while the brakes were applied were 280 ft long, and the tread on the tires produced a coefficient of kinetic friction of 0.30 with the road. (a) In your testimony in court, will you say that the driver was obeying the posted speed limit? You must be able to back up your answer with clear numerical reasoning during cross-examination. (b) If the driver's speeding ticket is \(\$ 10\) for each mile per hour he was driving above the posted speed limit, would he have to pay a ticket, and if so, how much would it be?

\(\bullet\) \(\bullet\) A 2.50 -kg mass is pushed against a horizontal spring of force constant 25.0 \(\mathrm{N} / \mathrm{cm}\) on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store 11.5 \(\mathrm{J}\) of potential energy in it, the mass is sud- denly released from rest. (a) Find the greatest speed the mass reaches. When does this occur? (b) What is the greatest accel- eration of the mass, and when does it occur?

\(\cdot\) A 0.145 kg baseball leaves a pitcher's hand at a speed of 32.0 \(\mathrm{m} / \mathrm{s} .\) If air drag is negligible, how much work has the pitcher done on the ball by throwing it?
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