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Chapter 7: Problem 10

\(\cdot\) \(\cdot\) A tow truck pulls a car 5.00 \(\mathrm{km}\) along a horizontal road- way 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 for horizontal, 3480750 J for 35 degrees. (b) Same as (a). (c) Gravity does 0 J work.

Step by step solution

01

Understand the Formula for Work

Work done by a force is 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 moved in the direction of the force, and \( \theta \) is the angle between the force and the direction of movement.
02

Calculate Work for Horizontal Pull

In the first case, the cable pulls horizontally, meaning \( \theta = 0^\circ \). Using \( W = F \cdot d \cdot \cos(\theta) \), substitute \( F = 850 \text{ N} \), \( d = 5000 \text{ m} \) (5 km) and \( \theta = 0 \). Thus, \( W = 850 \cdot 5000 \cdot \cos(0) = 4250000 \text{ J} \).
03

Calculate Work for Pull at 35 Degrees

Now consider when the cable pulls at \( 35.0^\circ \) above the horizontal. Here, \( \theta = 35.0^\circ \). The formula becomes \( W = 850 \cdot 5000 \cdot \cos(35^\circ) \). Calculate \( \cos(35^\circ) \approx 0.819 \), so \( W \approx 850 \cdot 5000 \cdot 0.819 = 3480750 \text{ J} \).
04

Work Done on the Tow Truck

Whether the cable pulls horizontally or at an angle, the work done on the tow truck is essentially internal to the truck and involves friction and mechanical work within the vehicle's systems. For simplicity and the context provided, assume the same amount of work is done by friction forces as done on the car for towing purposes. Thus, the work calculated in parts (a) applies here for movement influenced by the tension: 4250000 J when horizontal and 3480750 J at 35 degrees.
05

Work Done by Gravity

When pulling horizontally, gravity does no work because the displacement in the horizontal direction doesn't have a component in the direction of the gravitational force (vertical). Therefore, the work done by gravity in part (a) is 0 J.

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Key Concepts

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

Work Calculation
Calculating the amount of work done is a fundamental concept in physics, especially when it comes to moving objects. The work done by a force can be determined using the formula: \[ W = F \cdot d \cdot \cos(\theta) \] where:
  • \( W \) is the work done in joules (J).
  • \( F \) is the force applied in newtons (N).
  • \( d \) is the distance in meters (m) over which the force is applied.
  • \( \theta \) is the angle between the force and the direction of movement.
This formula highlights that work is the component of the force that acts in the direction of movement multiplied by the distance.
In situations where the angle \( \theta \) is zero, the entire force contributes to the work. When the angle is greater than zero, only a part of the force does work, as shown by the cosine factor.
Forces and Angles
Understanding how forces act at angles is crucial in accurately calculating work. When we have a force acting at an angle, such as the tension in a cable pulling a car, only the component of the force that acts along the direction of movement does work.
To find this component, we use the cosine of the angle \( \theta \).
  • If \( \theta = 0^\circ \), the force is fully in the direction of movement, so \( \cos(\theta) = 1 \) and the work done is maximal.
  • For an angle \( \theta = 35^\circ \), the work reduced since \( \cos(35^\circ) \approx 0.819 \).
This reduction in work illustrates that as the angle increases from zero, less of the force is used effectively for doing work along the desired direction. Calculating these angles ensures we understand the real influence of the force.
Gravitational Work
Gravitational work refers to the work done by the gravitational force. In scenarios like pulling a car on a horizontal surface, gravity exerts a force perpendicular to the movement.
Consequently, it does not do any work on the car as work is only done when there is displacement in the direction of the force. For horizontal movement:
  • The gravitational force acts vertically.
  • The car moves horizontally, so the displacement has no component in the vertical direction.
As a result, the gravitational work is zero Joules. This concept is significant when analyzing forces in physics, ensuring we only account for forces working parallel to the direction of movement.
Physics Problem Solving
When tackling physics problems like this one, a systematic problem-solving approach is key. The process usually follows these steps:
  • Identify the known values and the unknowns. For example, the force (850 N), distance (5000 m), and angles (0掳 and 35掳).
  • Apply relevant formulas, such as the work calculation formula \( W = F \cdot d \cdot \cos(\theta) \).
  • Perform calculations carefully, ensuring to convert units if necessary, such as kilometers to meters.
  • Consider the direction of forces and movement to determine which forces do work.
Analyzing each factor, like forces and angles, allows solvers to pinpoint the contribution of each to the overall problem. This strategy not only simplifies complex problems but also sharpens analytical and calculation skills in physics.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) Human energy vs. insect energy. For its size, the com- mon flea is one of the most accomplished jumpers in the animal world. A \(2.0-\mathrm{mm}\) -long, 0.50 \(\mathrm{mg}\) critter can reach a height of 20 \(\mathrm{cm}\) in a single leap. (a) Neglecting air drag, what is the take- off speed of such a flea? (b) Calculate the kinetic energy of this flea at takeoff and its kinetic energy per kilogram of mass. (c) If a 65 kg, 2.0 -m-tall human could jump to the same height com- pared with his length as the flea jumps compared with its length, how high could he jump, and what takeoff speed would he need? (d) In fact, most humans can jump no more than 60 \(\mathrm{cm}\) from a crouched start. What is the kinetic energy per kilogram of mass at takeoff for such a 65 kg person? (e) Where does the flea store the energy that allows it to make such a sudden leap?

\(\cdot\) Meteor crater. About \(50,000\) years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Recent \((2005)\) measurements estimate that this meteor had a mass of about \(1.4 \times 10^{8} \mathrm{kg}\) (around \(150,000\) tons) and hit the ground at 12 \(\mathrm{km} / \mathrm{s}\) . (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy produced in one day by a standard coal-fired power plant, which generates about 1 billion joules per second?

\(\bullet\) Tendons. Tendons are strong elastic fibers that attach mus- cles to bones. To a reasonable approximation, they obey Hooke's law. In laboratory tests on a particular tendon, it was found that, when a 250 g object was hung from it, the tendon stretched 1.23 \(\mathrm{cm}\) . (a) Find the force constant of this tendon in \(\mathrm{N} / \mathrm{m}\) . (b) Because of its thickness, the maximum tension this tendon can support without rupturing is 138 \(\mathrm{N}\) . By how much can the tendon stretch without rupturing, and how much energy is stored in it at that point?

\(\bullet\) \(\bullet\) 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?

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