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Using the information for #1...
Using the information for #1 at http://gaberosenbaum.com/Champlain/NYA/2011_Winter/Excercises/NYA_PS_Work_Energy.pdf determine the work and motion of the 10.0 kg object.
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For simple mechanical systems, work can be defined as the force applied times the distance an object travels in the direction of the force. or W = Fdcos(theta)
If we assume from the information on the assigned web site that the object and force are co-directional, we can determine the amount of work by calculating the area under the line.
W = 40NX1m + 10NX4m + 0NX4m = 80 joules.
So, the total work is 40 joules.
According to the work energy theorem, the change in total mechanical energy of the object will be equal to the work applied. If we again assume that the object is not changing height due to the work, then the entire amount of work will go into changing the object's kinetic energy.
W = KEf - KEi. Because the initial speed is 0, the initial kinetic energy = 0, thus the work will be equal to the final kinetic energy
KE = W
1/2 MV^2 = W
V^2 = 2W/M
V = sqrt(2W/M) = sqrt(80/10) = sqrt(8) = 2.83 m/s
Posted by mwmovr40 on March 29, 2012 at 9:15 AM (Answer #1)
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