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Chapter 6: Problem 6

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of \(1.80 \times 10^{6} \mathrm{N}\) , one \(14^{\circ}\) west of north and the other \(14^{\circ}\) east of north, as they pull the tanker 0.75 \(\mathrm{km}\) toward the north. What is the total work they do on the supertanker?

Short Answer

Expert verified
The total work done is \( 2.62 \times 10^9 \: \text{Joules} \).

Step by step solution

01

Calculate the Work Done by Each Tugboat

The work done by a force can be calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( W \) is the work done, \( F \) is the force, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and the direction of motion. Both tugboats exert the same force of \( 1.80 \times 10^6 \; N \) over a distance of \( 0.75 \; km = 750 \; m \), and each at an angle of \( 14^{\circ} \) from north. As they pull towards the north, the angle used in the work calculation is \( 14^{\circ} \). Thus, the work done by each tugboat is \( W = 1.80 \times 10^6 \times 750 \times \cos(14^{\circ}) \).
02

Solve for the Work Done by Each Tugboat

Calculate the work done by each tugboat using the previously outlined formula: \[ W = 1.80 \times 10^6 \times 750 \times \cos(14^{\circ}) \]Using a calculator, compute \( \cos(14^{\circ}) \approx 0.9703 \). Then the work done by one tugboat is \[ W \approx 1.80 \times 10^6 \times 750 \times 0.9703 \approx 1.31 \times 10^9 \: \text{Joules} \]
03

Calculate the Total Work Done by Both Tugboats

Since both tugboats perform the same amount of work, the total work done is simply twice the work done by one. Therefore, \[ \text{Total Work (W)} = 2 \times 1.31 \times 10^9 = 2.62 \times 10^9 \: \text{Joules} \]
04

Conclusion

Thus, the total work done by both tugboats in pulling the supertanker is \( 2.62 \times 10^9 \: \text{Joules} \).

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Key Concepts

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

Vector Components
When forces are applied at an angle, it's often useful to break them down into their components. A force vector can be split into horizontal and vertical components. In the case of the tugboats pulling the supertanker, the force vectors are at angles to the north. This means each tugboat's force can be divided into:
  • A component pointing north (the direction of motion).
  • A component pointing east or west (perpendicular to the direction of motion).

To focus on how much of the force contributes to the actual movement of the tanker, we use the trigonometric function cosine. The component of the force doing work on the tanker is the portion of the force aligned with the northward motion. Thus, the northward component is calculated using:
  • \( F_{north} = F \cdot \cos(\theta) \)
By resolving the vectors, we ensure that when calculating work, we only consider the effective force moving the object in the desired direction.
Force Calculations
In physics, force calculations help us understand how interactions between objects result in changes in motion. The given problem involves calculating the force exerted by each tugboat. Each tugboat applies a constant force of:
  • \( F = 1.80 \times 10^6 \; \text{N} \)
Since the force is applied at an angle of 14 degrees to the north, only part of this force contributes to the motion of the tanker. The effective force used in the work calculation is found using:
  • \( F \, \cos(\theta) \), with \( \theta \) being 14 degrees.
This calculation not only determines the usable part of the force but also ensures that the tanker is moved efficiently towards the north, aligning with the desired direction of travel.
Work Done by a Force
To quantify how much energy a force transfers to an object, we calculate the work done. Work is computed using the formula:
  • \( W = F \cdot d \cdot \cos(\theta) \)
where \( F \) is the force applied, \( d \) is the distance over which the force acts, and \( \theta \) is the angle between the force and the direction of motion.
In the scenario with the tugboats, they exert their force over a distance of 750 meters north. This distance, combined with the force's northward component due to the angle, gives us the work each tugboat does:
  • \( W = 1.80 \times 10^6 \times 750 \times \cos(14^{\circ}) \)
After calculating the work for one tugboat, we double this value to find the total work done by both, leading to the total work of \( 2.62 \times 10^9 \) Joules. This reflects the combined energy transferred to move the supertanker in the specified direction.

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Most popular questions from this chapter

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of \(40.0^{\circ}\) above the horizontal. The glider has mass 0.0900 \(\mathrm{kg}\) . The spring has \(k=640 \mathrm{N} / \mathrm{m}\) and negligible mass. When the spring is released, the glider travels a maximum distance of 1.80 \(\mathrm{m}\) m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. (a) What distance was the spring originally compressed? (b) When the glider has traveled along the air track 0.80 \(\mathrm{m}\) from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

Springs in Parallel. Two springs are in parallel if they are parallel to each other and are connected at their ends (Figure 6.33\()\) . We can think of this combination as being equivalent to a single spring. The force constant of the equivalent single spring is called the effective forceconstant, \(k_{\text { eff }},\) of the combination. (a) Show that the effective force constant of this combination is \(k_{\mathrm{eff}}=k_{1}+k_{2} .\) (b) Generalize this result for \(N\) springs in parallel.

Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 \(\mathrm{J}\) of work when you compress the springs 0.200 \(\mathrm{m}\) from their uncompressed length. (a) What magnitude of force must you apply to hold the platform in this position? (b) How much additional work must you do to move the platform 0.200 \(\mathrm{m}\) farther, and what maximum force must you apply?

A physics professor is pushed up a ramp inclined upward at \(30.0^{\circ}\) above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 85.0 \(\mathrm{kg}\) . He is pushed 2.50 \(\mathrm{m}\) along the incline by a group of students who together exert a constant horizontal force of 600 \(\mathrm{N}\) . The professor's speed at the bottom of the ramp is 2.00 \(\mathrm{m} / \mathrm{s}\) . Use the work-energy theorem to find his speed at the top of the ramp.

A little red wagon with mass 7.00 \(\mathrm{kg}\) moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 \(\mathrm{m} / \mathrm{s}\) and then is pushed 3.0 \(\mathrm{m}\) in the direction of the initial velocity by a force with a magnitude of 10.0 \(\mathrm{N}\) . (a) Use the work- energy theorem to calculate the wagon's final speed. (b) Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships of Chapter 2 to calculate the wagon's final speed. Compare this result to that calculated in part (a).
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