imported>Patrick: Created page with ''''(a)''' The diagram below shows the forces acting on the mass and a free body diagram. The displacement of the mass is $\Delta x = x$ because we chose the initial po…'

2011-08-12T16:30:43Z

Created page with ''''(a)''' The diagram below shows the forces acting on the mass and a free body diagram. The displacement of the mass is <math>\Delta x = x</math> because we chose the initial po…'

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'''(a)''' The diagram below shows the forces acting on the mass and a free body diagram. The displacement of the mass is <math>\Delta x = x</math> because we chose the initial position of the mass as the origin of the coordinate system (<math>x_i = 0</math> and <math>x_f = x</math>). 
[[image:Helena_Work-Energy_Ex_9_SolnA.png|TOP]] 

#The work done by the normal force is zero because it is perpendicular to the displacement or because the x-component of the normal force is zero. 
#The gravitational force is a constant force. The x-component of the gravitational force is <math>{F_G}_x = F_G cos(-53^{\circ}) = (10)(0.6) = 6 N</math> and so we may use <math>W_{F_G} = {F_G}_x \Delta x = 6 x J</math>. We cannot compute its value because we don't know the value of x. 
#Before we compute the work done on the mass by friction we have to first determine the magnitude of the force of friction. Applying the Second Law to the y-components of forces we find that the normal force is 8 N and that, given the coefficient of friction is <math>\mu_k = 0.1</math>, the force of friction is <math>F_f = (0.1)(8) = 0.8 N</math>. The x-component of the frictional force is <math>{F_f}_x = F_f \cos(180^{\circ}) = -0.8 N</math>. From there we find the work done by friction <math>W_{F_f} = {F_f}_x \Delta x = -0.8 x J</math>. Again, we cannot compute the value of this force because we don't know the value of x. 
#The force exerted on the mass by the spring is not constant. We have to use a graph to find it. As the block slides down the x-component of the spring force is <math>{F_{sp}}_x = F_{sp} cos(180^{\circ}) = -F_{sp}</math>. Note that on the vertical axis we are plotting the x-component of the spring force which is negative (the graph is below the axis). 
[[image:Helena_Work-Energy_Ex_9_SolnB.png|TOP]] 

The area under the graph is <math>(\frac{1}{2} x) \times (-k x) = -\frac{1}{2} k x^2 = -\frac{1}{2} 250 x^2 = -125 x^2 J</math>. Therefore <math>W_{F_{sp}} = -125 x^2 J</math>. 

'''(b)''' The net work is '''zero''' because the mass starts from rest and stops when it slides down the distance <math>x</math>. The change of the kinetic energy is zero. 

'''(c)''' Knowing that <math>W_{net} = 0 = W_{F_G} + W_{F_f} + W_{F_{sp}}</math> we can compute <math>x</math>. By substituting the expressions found above into this equation we write:

<math>0 = 6 x - 0.8 x - 125 x^2</math>
 
Solving for <math>x</math> we find

<math>x = 0.0416 = 4.16 cm</math>.

Work-Energy Theorem EX 9 - Revision history

imported>Patrick: Created page with ''''(a)''' The diagram below shows the forces acting on the mass and a free body diagram. The displacement of the mass is \Delta x = x because we chose the initial po…'

imported>Patrick: Created page with ''''(a)''' The diagram below shows the forces acting on the mass and a free body diagram. The displacement of the mass is $\Delta x = x$ because we chose the initial po…'