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Chapter 7: 36P (page 173)

A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in Fig.7-39. The scale of the figure’s vertical axis is set byFs=10.0N. How much work is done by the force as the block moves from the origin to x=8.0cm?

Short Answer

Expert verified

The net work done by the force using a graph W=25J.

Step by step solution

01

Given

  1. The mass of block is, m=5.0kg
  2. The scale for vertical axis is, Fs=10.0N
02

Understanding the concept

We can use the concept that the work done is equal to the area under the curve for a graph.

Formula:

W=∫xixfF(x)dxA=12×base×heightA=length×breadth

03

Calculate the net work done

We know that the total area will be the net work done by the force.

So,

W=W1+W2+W3+W4

Substitute all the value in the above equation.

W=10N×2m+12×10N×2m+0+-12×2N×5mW=25J

Therefore, the net work done is 25J

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Most popular questions from this chapter

Figure 7 -42 shows a cold package of hot dogs sliding rightward across a frictionless floor through a distance d=20.0cmwhile three forces act on the package. Two of them are horizontal and have the magnitudes F1=5.00NandF2=1.00N; the third is angled down byθ=60.0° and has the magnitudeF3=4.00N. (a) For the 20.0 cm displacement, what is the net work done on the package by the three applied forces, the gravitational force on the package, and the normal force on the package? (b) If the package has a mass of 2.0 Kg and an initial kinetic energy of 0, what is its speed at the end of the displacement?

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