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Chapter 12: Problem 50

A tow truck drags a stalled car along a road. The chain makes an angle of \(30^{\circ}\) with the road and the tension in the chain is \(1500 \mathrm{N}\). How much work is done by the truck in pulling the car \(1 \mathrm{km}\) ?

Short Answer

Expert verified
The work done by the truck is approximately 1,299,000 joules.

Step by step solution

01

Understand the Work Formula

The formula to calculate work is \( W = F \, \cdot \, d \, \cdot \, \cos(\theta) \), where \( W \) is the work done, \( F \) is the force applied, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and the direction of movement.
02

Identify Known Values

From the problem, we know:- The force \( F = 1500 \, \mathrm{N} \),- The distance \( d = 1 \, \mathrm{km} = 1000 \, \mathrm{m} \),- The angle \( \theta = 30^{\circ} \).
03

Calculate the Work Done

Substitute the known values into the work formula:\[W = 1500 \, \mathrm{N} \, \times \, 1000 \, \mathrm{m} \, \times \, \cos(30^{\circ})\]The cosine of \( 30^{\circ} \) is \( \frac{\sqrt{3}}{2} \).
04

Perform the Calculation

Calculate \( \cos(30^{\circ}) \) which is \( \frac{\sqrt{3}}{2} \) and substitute back into the equation:\[W = 1500 \, \times \, 1000 \, \times \, \frac{\sqrt{3}}{2}\]Calculate:\[W = 1500 \, \times \, 1000 \, \times \, 0.866 = 1299000 \, \mathrm{J} \text{ (approximately)}\]
05

Unit Verification

The units for the calculated work are in joules since we used newtons for force and meters for distance, confirming that the work is accurately calculated as \( 1299000 \, \mathrm{J} \).

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Key Concepts

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

Work Done
When it comes to the concept of work done, think of it as the energy transferred to an object when a force is applied while the object moves some distance. The amount of work done depends on three key factors: the magnitude of the force applied, the distance moved by the object, and the angle at which the force is applied relative to the displacement.

The formula that encapsulates this concept is given by: \[ W = F \cdot d \cdot \cos(\theta) \] where:
  • \(W\) is the work done in joules (J).
  • \(F\) is the force applied, measured in newtons (N).
  • \(d\) is the distance traveled by the object in meters (m).
  • \(\theta\) is the angle between the force direction and the displacement direction.
From this equation, you can see how work is directly proportional to both the force exerted and the distance the object moves when the force and movement are in the same direction.
Force and Distance
In physics, force and distance are two crucial elements that contribute directly to the amount of work done on an object. Let's delve into each component:

  • **Force**: It is a push or a pull exerted on an object, measured in newtons (N). The force must be actively applied and is often constant in problems involving work done.
  • **Distance**: This is the measure of how far the object moves while the force is applied, usually converted into meters in calculations. Remember, only the distance moved in the direction of the force counts towards work done.
An essential thing to remember is that if the distance across which a force is applied is zero, even if some force is exerted, the work done is zero because the object hasn't moved.
Angle of Force
In calculating work done, the angle at which a force is applied relative to the direction of movement impacts the result significantly. This angle, denoted as \( \theta \), determines how effectively the force contributes to work.

  • If \(\theta = 0^{\circ}\): The entire force contributes to work done because the force direction aligns perfectly with the direction of movement.
  • If \(\theta = 90^{\circ}\): The force does no work because it acts perpendicular to the direction of movement, contributing zero to the distance moved in that direction.
  • For all other angles, use \(\cos(\theta)\) to find the effective component of the force used in calculating work done.
In practical scenarios, such as our tow truck example, the \(\cos(30^{\circ})\) helps determine how much of the force actually propels the car forward, thus influencing the final work done outcome.

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