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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) 4250000 J horizontally, 3480750 J at 35掳; (b) equal work on the truck; (c) 0 J by gravity.

Step by step solution

01

Understanding the Problem

We are given a scenario where a tow truck is pulling a car with a cable, and we need to calculate work done by the cable and gravity under different conditions. The work done is given by the formula \( W = F \cdot d \cdot \cos(\theta) \) where \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force and direction of motion.
02

Work Done on the Car: Cable Pulling Horizontally

When the cable pulls horizontally, the angle \( \theta = 0^{\circ} \). The work done is calculated as \[ W = F \cdot d \cdot \cos(0^{\circ}) = 850 \, \mathrm{N} \times 5.00 \, \mathrm{km} \times \cos(0^{\circ}) = 850 \, \mathrm{N} \times 5000 \, \mathrm{m} = 4250000 \, \mathrm{J}. \]
03

Work Done on the Car: Cable Pulling at an Angle

When the cable pulls at \( 35.0^{\circ} \) above the horizontal, use the work formula: \[ W = F \cdot d \cdot \cos(35.0^{\circ}) = 850 \, \mathrm{N} \times 5000 \, \mathrm{m} \times \cos(35.0^{\circ}). \] Calculate \( \cos(35.0^{\circ}) \approx 0.819 \):\[ W = 850 \, \mathrm{N} \times 5000 \, \mathrm{m} \times 0.819 = 3480750 \, \mathrm{J}. \]
04

Work Done on the Tow Truck

For both horizontal and \( 35.0^{\circ} \) pulls, the situation with the tow truck is similar to the car since the same force applies, just working the opposite way. The work done on the tow truck is the same amount but opposite signs for each scenario respectively.
05

Work Done by Gravity on the Car

Gravity acts perpendicular to the direction of motion (assuming no vertical displacement), so the angle here \( \theta = 90^{\circ} \). Therefore, \[ W = F \cdot d \cdot \cos(90^{\circ}) = 0 \, \mathrm{J}. \]No work is done by gravity in the horizontal movement of the car.

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Key Concepts

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

Physics Problem Solving
Physics problem solving is a structured approach to understanding and finding solutions to real-world situations involving physical principles. It involves breaking down a problem into smaller, manageable steps and applying physics concepts to solve each part. This method often starts with a clear understanding of the given scenario. For instance, in the tow truck exercise, we begin by identifying the force applied, the distance moved, and the angle at which the force is applied.

  • The main formula used in these problems is the work formula: \( W = F \cdot d \cdot \cos(\theta) \).
  • Here, \( W \) represents work, \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force direction and motion.
Once you have these parameters identified, the solution often involves plugging them into the equation and performing basic arithmetic operations to arrive at an answer. Problem solving in physics requires a firm grasp of concepts and a logical, step-by-step approach.
Forces and Motion
Understanding forces and motion is crucial in physics as they govern how objects interact with each other. In our scenario, the force involved is the tension from the tow truck's cable. This force is what moves the car over a distance.

  • Force: Measured in Newtons \( (N) \), describes the interaction that causes an object to change either its speed or direction of motion.
  • Motion: Describes how the object moves when a force is applied.
The concept of motion is crucial as it determines the direction in which the work is done. Forces can act in different directions, and their component along the line of motion is what does work. In this exercise's context, we examined different angles. If the force acts horizontally, more of it contributes to moving the car forward. At angles (like \(35.0^\circ\)), only part of the force contributes to forward motion, slightly reducing the work done.
Work Calculation
The calculation of work in physics involves understanding how force causes movement over a distance. It requires applying the work formula, \( W = F \cdot d \cdot \cos(\theta) \), to quantify the energy transfer. Work is measured in Joules \( (J) \), which indicates the amount of energy spent to perform a task.

  • For a horizontal pull (\( \theta = 0^\circ \)), \( \cos(0^\circ) = 1 \), resulting in maximal work calculated as \( 4250000 \, J \).
  • For an angled pull (\( \theta = 35.0^\circ \)), \( \cos(35.0^\circ) \approx 0.819 \), resulting in reduced work compared to horizontal pull, \( 3480750 \, J \).
The work calculated reflects how much of the force contributes to the car's motion. Notice that gravity, acting perpendicularly to the car's movement, doesn't do work in this case. Work is only done by components of force in the direction of displacement, helping students understand the efficiency of energy transfer in relation to force and its direction.

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Most popular questions from this chapter

Cycling. For a touring bicyclist the drag coefficient \(C\left(f_{\text { air }}=\frac{1}{2} C A \rho v^{2}\right)\) is \(1.00,\) the frontal area \(A\) is \(0.463 \mathrm{m}^{2},\) and the coefficient of rolling friction is \(0.0045 .\) The rider has mass 50.0 \(\mathrm{kg}\) , and her bike has mass 12.0 \(\mathrm{kg}\) . (a) To maintain a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) (about 27 \(\mathrm{mi} \mathrm{h}\) ) on a level road, what must the rider's power output to the rear wheel be? (b) For racing, the same rider uses a different bike with coefficient of rolling friction 0.0030 and mass 9.00 \(\mathrm{kg}\) . She also crouches down, reducing her drag coefficient to 0.88 and reducing her frontal area to 0.366 \(\mathrm{m}^{2} .\) What must her power output to the rear wheel be then to maintain a speed of 12.0 \(\mathrm{m} / \mathrm{s} ?\) (c) For the situation in part \((\mathrm{b}),\) what power output is required to maintain a speed of 6.0 \(\mathrm{m} / \mathrm{s}\) ? Note the great drop in power requirement when the speed is only halved. (For more on aerodynamic speed limitations for a wide variety of human-powered vehicles, see "The Aerodynamics of Human-Powered Land Vehicles," Scientific American, December 1983.

Crash Barrier. A student proposes a design for an auto- mobile crash barrier in which a \(1700-\mathrm{kg}\) sport utility vehicle moving at 20.0 \(\mathrm{m} / \mathrm{s}\) crashes into a spring of negligible mass that slows it to a stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater than 5.00\(g\) . (a) Find the required spring constant \(k\) , and find the distance the spring will compress in slowing the vehicle to a stop. In your calculation, disregard any deformation or crumpling of the vehicle and the friction between the vehicle and the ground. (b) What disadvantages are there to this design?

A sled with mass 8.00 \(\mathrm{kg}\) moves in a straight line on a frictionless horizontal surface. At one point in its path, its speed is 4.00 \(\mathrm{m} / \mathrm{s}\) : after it has traveled 2.50 \(\mathrm{m}\) beyond this point, its speed is 6.00 \(\mathrm{m} / \mathrm{s}\) . Use the work-energy theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sled's motion.

A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying, or resting; during these activities, she expends energy at an average rate of 100 W. If she expends a total of \(1.1 \times 10^{7} \mathrm{J}\) of energy in a 24 -bour day, how much of the day did she spend walking?

A luggage handler pulls a \(20.0-\mathrm{kg}\) suitcase up a ramp inclined at \(25.0^{\circ}\) above the horizontal by a force \(\overrightarrow{\boldsymbol{F}}\) of magnitude 140 \(\mathrm{N}\) that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is \(\mu_{\mathrm{k}}=0.300\) . If the suitcase travels 3.80 \(\mathrm{m}\) along the ramp, calculate (a) the work done on the suitcase by the force \(\overrightarrow{\boldsymbol{F}},\) (b) the work done on the suitcase by the gravitational force; (c) the work done on the suitcase by the normal force; (d) the work done on the suitcase by the friction force; (c) the total work done on the suitcase. (f) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 3.80 \(\mathrm{m}\) m along the ramp?
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