Mechanical work. Question 1

Question.

A 54.4-kg box is being pushed a distance of 8.27 m across the floor by a force F whose magnitude is 163 N. The force F is parallel to the displacement of the box. The coefficient of kinetic friction is 0.227. Determine the work done on the box by (a) the applied force, (b) the friction force, (c) the normal force, and (d) by the force of gravity. Be sure to include the proper plus or minus sign for the work done by each force.

Source. Yahoo! Answers!

Solution.

a) Applied Force F=163 N
So W – Work, D – distance
W=F*D=163*8.27=1348.01 J

b) friction force – f

k – coefficient of kinetic friction
9= 9.81 m/s^2

f=k*m*g = 0.227*54.4*9.81=121.14 N

W=-f*D= -8.27*121.14=-1001.83 J

c) W of normal force always equal zero

d) work by the force of gravity equal zero too / reason 90 degree between displacement and gravity force

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