imported>Patrick: Created page with ''''(a)''' The work is equal to the area under the force-position graph. From symmetry, we can see that from 0 m to 6 m, the area under the graph below the x-axis is equal to the …'

2011-08-04T18:42:54Z

Created page with ''''(a)''' The work is equal to the area under the force-position graph. From symmetry, we can see that from 0 m to 6 m, the area under the graph below the x-axis is equal to the …'

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'''(a)''' The work is equal to the area under the force-position graph. From symmetry, we can see that from 0 m to 6 m, the area under the graph below the x-axis is equal to the area under the graph above the x-axis. We only need to calculate the area under the graph from 6 m to 10 m. The work done is

<math>W = 20 J + 10 J = 30 J</math>.
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[[image:Helena_Work-Energy_Ex_8_Soln.png|TOP]]<br><br>

'''(b)''' The change in kinetic energy is equal to the work done:

<math>\Delta K = W = 30 J</math>

<math>\frac{1}{2} m {v_2}^2 - \frac{1}{2} m {v_1}^2 = 30 J</math>

<math>\frac{1}{2} (2) {v_2}^2 - \frac{1}{2} (2) (2)^2 = 30 J</math>

<math>\frac{1}{2} (2) {v_2}^2 - 4 = 30 J</math>

<math>{v_2}^2 = {34 J \over 1 kg}</math>

<math>v_2 = 5.83 m/s</math>

Work-Energy Theorem EX 8 - Revision history

imported>Patrick: Created page with ''''(a)''' The work is equal to the area under the force-position graph. From symmetry, we can see that from 0 m to 6 m, the area under the graph below the x-axis is equal to the …'