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Chapter 6: Problem 2

A tow truck pulls a car 5.00 \(\mathrm{km}\) along a horizontal roadway using a cable having a tension of 850 \(\mathrm{N}\) . (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

Short Answer

Expert verified
(a) 4250 kJ, 3480.75 kJ; (b) -4250 kJ, -3480.75 kJ; (c) 0 J.

Step by step solution

01

Understand Work Calculation

Work done is computed with the formula: \[ W = F imes d imes \cos(\theta) \]where:- \(W\) is the work done,- \(F\) is the force applied (850 N in this case),- \(d\) is the distance over which the force is applied (5.00 km = 5000 m),- \(\theta\) is the angle between the force direction and the direction of movement.
02

Calculate Work for Horizontal Pull

When the cable pulls horizontally, \( \theta = 0^{\circ} \). Thus, the work done by the cable on the car is:\[ W = 850 \times 5000 \times \cos(0^{\circ}) \]\[ W = 850 \times 5000 \times 1 \]\[ W = 4250000 \, \mathrm{J} \]
03

Calculate Work for Pull at 35 Degrees

When the cable pulls at \(35^{\circ}\) above the horizontal, the work done is:\[ W = 850 \times 5000 \times \cos(35^{\circ}) \]Using \( \cos(35^{\circ}) \approx 0.819 \):\[ W = 850 \times 5000 \times 0.819 \]\[ W \approx 3480750 \, \mathrm{J} \]
04

Understand Work on Tow Truck

The work done on the tow truck by the cable will be equal in magnitude but opposite in direction (negative) since the movement is the same.
05

Calculate Work on Tow Truck for Horizontal and 35 Degrees

- For horizontal pull, the work done on the tow truck is: \[ W = - 4250000 \, \mathrm{J} \]- For pull at \(35^{\circ}\): \[ W = - 3480750 \, \mathrm{J} \]
06

Evaluate Work Done by Gravity

Gravity does no work since the motion is horizontal and perpendicular to the force of gravity. Thus, work done by gravity is 0 J.

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Key Concepts

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

Physics Problems
In physics, when dealing with problems like the one involving a tow truck and a car, it's essential to dissect the situation into manageable components. This approach helps in understanding the principles at play, such as forces and motion.
Physics problems often involve concepts like work and energy, which are central to understanding how force moves an object over a distance. The principle guiding these problems is the work-energy principle.
  • Work is the energy transferred by a force over a distance in the direction of the force.
  • Energy concepts help describe whether a task was physically taxing or required significant effort.
In our specific problem, we're analyzing the work done as a car is pulled horizontally by a tow truck. Understanding the way forces interact in such scenarios allows for the resolution of various physics challenges.
Force and Tension
Force is a vector quantity, which means it has both magnitude and direction. When a cable pulls a car, the tension in the cable is the force exerted. Tension is the pulling force transmitted axially by the string or cable.
  • Tension can be visualized as the cables being taut under a pulling force.
  • It's measured in newtons (N) and affects how work is calculated in our exercise.
In this scenario, the tension is 850 N, directly influencing the work done on the car. Understanding force and tension helps in predicting how objects will accelerate or resist the force applied, essential in solving physics problems.
Angle of Application
The angle of application refers to the angle at which a force is applied concerning a reference direction, often the horizontal surface in this context. The importance of the angle is shown in the equation for work: \[ W = F \times d \times \cos(\theta) \] The angle, \(\theta\), determines how much of the force effectively contributes to moving the object in the desired direction.
  • At \(\theta = 0^{\circ}\), all the force contributes to the movement.
  • At \(35^{\circ}\), only a part of the force does, as represented by the cosine of the angle.
This specificity affects the total work done and ultimately demonstrates how work depends not just on force and distance, but also on direction.
Horizontal Motion
Horizontal motion is straightforward when considering displacement along a flat plane. This motion is critical in the exercise, where a car is pulled by a truck along a horizontal road.
  • All forces acting vertically, like gravity, do not perform work in purely horizontal motion.
  • This scenario allows us to focus on the horizontal component of the force exerted.
By understanding horizontal motion, one can see that any vertical forces, like gravity in this situation, are perpendicular to the movement. As such, they do not contribute to the work done, simplifying our calculations drastically.

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Most popular questions from this chapter

CALC A net force along the \(x\) -axis that has \(x\) -component \(F_{x}=-12.0 \mathrm{N}+\left(0.300 \mathrm{N} / \mathrm{m}^{2}\right) x^{2}\) is applied to a 5.00 \(\mathrm{-kg}\) object that is initially at the origin and moving in the \(-x\) -direction with a speed of 6.00 \(\mathrm{m} / \mathrm{s} .\) What is the speed of the object when it reaches the point \(x=5.00 \mathrm{m} ?\)

A 2.50 -kg textbook is forced against a horizontal spring of negligible mass and force constant \(250 \mathrm{N} / \mathrm{m},\) compressing the spring a distance of 0.250 \(\mathrm{m} .\) When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.30 .\) Use the work-energy theorem to find how far the textbook moves from its initial position before coming to rest.

A ski tow operates on a \(15.0^{\circ}\) slope of length 300 \(\mathrm{m} .\) The rope moves at 12.0 \(\mathrm{km} / \mathrm{h}\) and provides power for 50 riders at one time, with an average mass per rider of 70.0 \(\mathrm{kg} .\) Estimate the power required to operate the tow.

A Near-Earth Asteroid. On April \(13,2029\) (Friday the 13th!), the asteroid 99942 Apophis will pass within \(18,600\) mi of the earth- about \(\frac{1}{13}\) the distance to the moon! It has a density of \(2600 \mathrm{kg} / \mathrm{m}^{3},\) can be modeled as a sphere 320 \(\mathrm{m}\) in diameter, and will be traveling at 12.6 \(\mathrm{km} / \mathrm{s}\) . (a) If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver? (b) The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases \(4.184 \times 10^{15}\) J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?

A tandem (two-person) bicycle team must overcome a force of 165 \(\mathrm{N}\) to maintain a speed of 9.00 \(\mathrm{m} / \mathrm{s} .\) Find the power required per rider, assuming that each contributes equally. Express your answer in watts and in horsepower.
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