Asked by Anonymous
3. You push your physics book 1.50 m along a horizontal table- top with a horizontal push of 2.40 N while the opposing force of friction is 0.600 N. How much work does each of the following forces do on the book:
(a) your 2.40-N push,
(b) the friction force,
(c) the normal force from the tabletop, and
(d) gravity?
(e) What is the net work done on the book?
Answers
Answered by
bobpursley
I will be happy to critique your thinking.
Answered by
Mirian
a) W=2.4*1.5*cos(0)= 3.6J
b) W=0.600*1.5*cos(180)= -0.9J
c) W=0J because the displacement and the normal force is perpendicular to each other so cos(90)=0
d) W=0J for the same reason as part (c), but with the gravity
e) W(net)= 3.6+(-0.9)+0+0= 2.7J
you are just adding all the works together
b) W=0.600*1.5*cos(180)= -0.9J
c) W=0J because the displacement and the normal force is perpendicular to each other so cos(90)=0
d) W=0J for the same reason as part (c), but with the gravity
e) W(net)= 3.6+(-0.9)+0+0= 2.7J
you are just adding all the works together
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