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Chapter 6: Problem 15

A crane lifts a 650 -kg beam vertically upward 23 m and then swings it eastward \(18 \mathrm{m}\). How much work does the crane do? Neglect friction, and assume the beam moves with constant speed.

Short Answer

Expert verified
The crane does 146,759.5 J of work on the beam.

Step by step solution

01

Calculate the force of gravity on the beam

First, calculate the force of gravity on the beam. The force due to gravity is given by \( F = m*g \) where \( m = 650 kg \) is the mass and \( g \approx 9.81 m/s^2 \) is the acceleration due to gravity. So, \( F = 650 kg * 9.81 m/s^2 = 6376.5 N \)
02

Identify the distance the beam is lifted

The beam is lifted vertical distance of \(23 m\). This is the distance over which the force is applied.
03

Calculation of the work done by the crane lifting

The work \(W\) done by the crane in lifting the beam is given by the equation \( W = F*d \) where \( F = 6376.5 N \) is the force and \( d = 23 m \) is the distance. So, \( W = 6376.5 N * 23 m = 146,759.5 J \)
04

Identify the work done by the crane swinging

The crane then swings the beam a different, horizontal distance of \(18 m\). However, because this motion is perpendicular to the direction of gravity, no additional work is done by the crane against the force of gravity while moving the beam sideways.
05

Calculate the total work done by the crane

The total work done by the crane is sum of work done in lifting and work done in swinging. Since no work is done in swinging, the total work is equal to work done in lifting the beam which is \( W = 146,759.5 J \)

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Most popular questions from this chapter

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