imported>Patrick at 06:22, 29 July 2011

2011-07-29T06:22:52Z

New page

[[image:Helena_Work_Ex_2_Soln.png|TOP]]<br><br>

{| border="1" cellpadding="2"
|-
!
! Displacement
! Work is equal to the area under the <math>F_x - x</math> graph
! Work
|-
! a)
| -4 to -2 m
| The area under the graph is <math>-0.5 \times 10 \times 2</math>
| -10 J
|-
! b)
| -4 to 0 m
| We use symmetry whenever possible to simplify the calculations. The area under the graph below the x-axis is equal to the area under the graph above the x-axis so the work done is zero.
| 0 J
|-
! c)
| 0 to 4 m
| From 2 to 4 m, the area under the graph below the x-axis is equal to the area under the graph above the x-axis so the work done is zero. We just have to calculate the area under the graph from 0 to 2 m: <math>10 \times 2</math>
| 20 J
|-
! d)
| -4 to 4 m
| Sum of areas
| 20 J
|-
! e)
| 0 to -2 m
| This problem is different from all the other shown in this example. The displacement in all the other cases was positive. In this case we move from the right to the left of the graph and therefore, the displacement is negative. Consequently, the area under the graph will be negative: <math>0.5 \times 10 \times (-2)</math>
| -10 J
|}

Work EX 2 - Revision history

imported>Patrick at 06:22, 29 July 2011