imported>Patrick at 16:34, 12 August 2011

2011-08-12T16:34:27Z

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'''(a)''' We choose the x-axis pointing to the right in the direction of the displacement. The x-component of the displacement is <math>\Delta x = 10 m</math> in this coordinate system. Then we draw a free body diagram which includes these forces: <math>F_G</math>, <math>F_N</math>, <math>F_R</math> and the force exerted by the man <math>F</math> - see the diagram below: 
[[image:Helena_Work-Energy_Ex_5_Soln.png|TOP]] 

Now we are ready to compute the work done by each force. The x-component of <math>F_G</math>, as well as the x-component of the normal force <math>F_N</math>, are zero and therefore the work done by these forces is equal to zero. Remember that whenever a force is perpendicular to the displacement the work done by this force is always zero.
 
The x-component of the force exerted by the man is <math>F_x = 80 \cos(-20^{\circ}) = 75.2 N</math>. The work done by this force is <math>W_F = (75.2)(10) = 752 J</math>.
 
The x-component of the resistive force exerted on the lawnmower is <math>{F_R}_x = 10 \cos(180^{\circ}) = -10 N</math>. The work done by this force is <math>W_{F_R} = (-10)(10) = -100 J</math>.
 
The total work done on the lawnmower is

<math>W_{total} = 752 + (-100) = 652 J</math>
 
'''(b)''' The sum of the y-components of forces is equal to zero. The sum of the x-components is equal <math>F_{net} = 75.2 - 10 = 65.2 N</math>. The work done by the net force is

<math>W_{total} = (65.2)(10) = 652 J</math>

We note that both methods yield the same result.

Work-Energy Theorem EX 5 - Revision history

imported>Patrick at 16:34, 12 August 2011