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Chapter 7: Problem 31

A constant 20-N force pushes a small ball in the direction of the force over a distance of \(5.0 \mathrm{m}\). What is the work done by the force?

Short Answer

Expert verified
The work done by the constant 20-N force over a distance of 5.0 m is \(W = 100 \, \text{J}\).

Step by step solution

01

Understand the formula for work done

Work done (W) is defined as the product of the force (F) and the distance (d) moved in the direction of the force. Mathematically, it is represented as: \( W = F \cdot d \cdot \cos{\theta} \) Where, W = Work done by the force, F = Magnitude of the force, d = Distance moved, 胃 = Angle between the force vector and the displacement vector. In this problem, since the force and displacement are in the same direction, the angle between them (胃) is 0 degrees (0 radians). The cosine of 0 degrees is 1.
02

Identify known quantities

Based on the problem statement, we can identify the following known quantities: - Force (F): 20 N - Distance (d): 5.0 m - Angle (胃): 0 degrees
03

Calculate the work done

Now we can substitute these values into the formula: \( W = F \cdot d \cdot \cos{\theta} \) \( W = (20 \, \text{N}) \cdot (5.0 \, \text{m}) \cdot \cos{0} \) Since \( \cos{0} = 1 \), the equation becomes: \( W = (20 \, \text{N}) \cdot (5.0 \, \text{m}) \) \( W = 100 \, \text{J} \) So, the work done by the force is 100 Joules.

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Key Concepts

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

Work-Energy Principle
The Work-Energy Principle is a fundamental concept in physics that connects the work done on an object with the change in its energy. When a constant force acts upon an object to move it, the energy of that object is altered. If the object moves in the direction of the force, kinetic energy is added to the object; conversely, if it moves against the force, the object loses kinetic energy.

In the context of our example with the small ball, the constant 20-N force doing work on the ball adds to the ball's kinetic energy. Here, the principle tells us that the amount of work done, which is 100 Joules, is the same as the increase in the ball's kinetic energy. So, effectively, we have transferred 100 Joules of energy from the force (often coming from a source like muscles or an engine) into the kinetic energy of the ball.
Force and Displacement
In physics, when we talk about work, we are usually referring to the work done when a force causes displacement. It's quite crucial to understand that for work to be done, the force must have a component in the direction of the displacement; otherwise, no work is done. This relationship is seen in the fact that work is a scalar quantity calculated as the product of the magnitude of the force, the displacement, and the cosine of the angle between the two vectors.

To delve deeper into our exercise situation, the force of 20 N is applied in the same direction as the displacement of 5.0 m. Therefore, this force results in positive work on the ball, pushing it along the 5.0 m path. If, however, the ball were to be pushed in a perpendicular direction to the force, no work would be done in the direction of the displacement, illustrating the importance of both magnitude and the direction of the force in work calculations.
Cosine of Angle in Work
The cosine of the angle in the work done equation is a mathematical representation of the direction component we just discussed. When the force and the displacement are parallel, like in our 20-N force acting over 5.0 m scenario, the angle between the force and the displacement () is zero degrees.Cosine of zero degrees is 1, which means the full magnitude of the force contributes to the work done.

If the force were applied at an angle, you would then need to use the cosine of that angle to find the component of the force that acts in the direction of the displacement. For example, if a force is acting at an angle of 60 degrees relative to the displacement, we would calculate (0.5) since (60 degrees). This principle underscores the influence of directional alignment between force and displacement in the physical work done on an object.

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Most popular questions from this chapter

How much work is done against the gravitational force on a 5.0 -kg briefcase when it is carried from the ground floor to the roof of the Empire State Building, a vertical climb of \(380 \mathrm{m} ?\)

Constant power \(P\) is delivered to a car of mass \(m\) by its engine. Show that if air resistance can be ignored, the distance covered in a time \(t\) by the car, starting from rest, is given by \(s=(8 P / 9 m)^{1 / 2} t^{3 / 2}\)

A 5.0-kg body has three times the kinetic energy of an 8.0 -kg body. Calculate the ratio of the speeds of these bodies.

Calculate the power output needed for a \(950-\mathrm{kg}\) car to climb a \(2.00^{\circ}\) slope at a constant \(30.0 \mathrm{m} / \mathrm{s}\) while encountering wind resistance and friction totaling \(600 \mathrm{N}\)

A bungee cord is essentially a very long rubber band that can stretch up to four times its unstretched length. However, its spring constant varies over its stretch [see Menz, P.G. "The Physics of Bungee Jumping." The Physics Teacher (November 1993) 31: 483-487]. Take the length of the cord to be along the \(x\) -direction and define the stretch \(x\) as the length of the cord \(l\) minus its un-stretched length \(l_{0} ;\) that is, \(x=l-l_{0}\) (see below). Suppose a particular bungee cord has a spring constant, for \(0 \leq x \leq 4.88 \mathrm{m},\) of \(k_{1}=204 \mathrm{N} / \mathrm{m}\) and for \(4.88 \mathrm{m} \leq x,\) of \(k_{2}=111 \mathrm{N} / \mathrm{m}\) (Recall that the spring constant is the slope of the force \(F(x) \text { versus its stretch } x .)\) (a) What is the tension in the cord when the stretch is \(16.7 \mathrm{m}\) (the maximum desired for a given jump)? (b) How much work must be done against the elastic force of the bungee cord to stretch it \(16.7 \mathrm{m} ?\)
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