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Chapter 6: Problem 2

Must you do work to whirl a ball around on the end of a string? Explain.

Short Answer

Expert verified
No, you must not do work to whirl a ball around on the end of a string. This is because the force applied and the displacement of the ball are perpendicular to each other, in which case the work done is zero.

Step by step solution

01

Define the Concept of Work

In physics, work is defined as the product of the force (F) applied on an object and the displacement (d) of the object in the direction of the force. Mathematically, it is given by the formula \(W = F \cdot d \cdot cos(胃)\), where \(胃\) is the angle between the direction of force and displacement.
02

Understand the Situation of the Ball

When whirling the ball around on a string, the force exerted on the ball is directed towards the center of the circle (this is the tension of the string), which causes the ball to move in a circular path. However, the ball's displacement at any moment is tangent to the circle, meaning it is perpendicular to the direction of the force.
03

Apply the Concept of Work to the Situation

Since the angle \(胃\) between the force and the displacement is 90 degrees (as they are perpendicular), and the cosine of 90 degrees is 0, substituting into the work formula results in the work done as 0 (since anything multiplied by 0 is zero). Thus, no work is done to whirl the ball.

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Most popular questions from this chapter

You"re an engineer for a company that makes bungee-jump cords, and you're asked to develep a formula for the work involved in stretching cords to double their length. Your cords have forcedistance relations described by \(F=-\left(k x+b x^{2}+c x^{3}+d x^{4}\right)\), where \(k, b, c\), and \(d\) are constants. (a) Given a cord with un stretched length \(L_{0}\). what's your formula? (b) Evaluate the work done in doubling the stretch of a 10 -m cord with \(k=420 \mathrm{~N} / \mathrm{m}\), \(b=-86 \mathrm{~N} / \mathrm{m}^{2}, c=12 \mathrm{~N} / \mathrm{m}^{3}\), and \(d=-0.50 \mathrm{~N} / \mathrm{m}^{4} .\)

If the scalar product of two nonzero vectors is zero, what can you conclude about their relative directions?

A force pointing in the \(x\)-direction is given by \(F=a x^{3 / 2}\), where \(a\) is a constant. The force does \(1.86 \mathrm{~kJ}\) of work on an object as the object moves from \(x=0\) to \(x=18.5 \mathrm{~m}\). Find the constant \(a\)

Show that the scalar product obeys the distributive law: \(\vec{A} \cdot(\vec{B}+\vec{C})=\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{C}\).

A force \(\vec{F}\) acts in the \(x\)-direction, its magnitude given by \(F=a x^{2}\), where \(x\) is in meters and \(a=3.0 \mathrm{~N} / \mathrm{m}^{2}\). Find the work done by this force as it acts on a particle moving from \(x=0\) to \(x=6.0 \mathrm{~m}\).
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