52 kg box pushed 7 m acoss floor by force p whos magnitude is 163N... the force is parallel to the displacement of the box. the coefficient of kinetic friction is .25. determine the work done on box by each of 4 forces that act on box. Include proper plus or minus sign.
a/. applied force.
b. frictional force.
c. Normal force
d. gravity
I got all right except for frictional force how do you find this? why is it not juts the negative of applied force? thank you
2007-02-18 00:33:17 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The frictional force acting is
µ mg.
0.25 x 52 x 9.8 N.
= 127.4 N
Work done by frictional force is force x distance moved.
0.25 x 52 x 9.8 x 7 J
=891.8 J.
Work done by the applied force is 163 x 7 =1141 J.
Out of this 1141 Joule 891.8 J were wasted against friction.
The balance 249.2 J is used for the work of moving the object through 7m.
The net force acting on the body is 163 - 127.4 = 35.6N.
Work done by this net force is 35.6 x 7 = 249.2 J
No work is done by the normal force or gravity as there is no displacement in their direction.
2007-02-18 03:19:57 · answer #1 · answered by Pearlsawme 7 · 0⤊ 0⤋
If I recall correcly the frictional force is the normal (in this case, weight) * coefficient of friction. You have the mass so working out the weight is easy.
If friction were always equal to the negative of the applied force, then you'd never be able to accelerate anything! It IS equal if the box is moving at uniform velocity - but you're not given that in the question.
2007-02-18 00:42:33 · answer #2 · answered by Anonymous · 0⤊ 0⤋
hello!!
if the frictional force was the opposite of the applied force the box would never move!!!
2007-02-18 01:09:53 · answer #3 · answered by Patrick N 2 · 0⤊ 0⤋
The frictional force is equals to
F=0.25x520
=130N
So resultant force
=163-130
=33N
Thus work done
=33x7
=231N
2007-02-18 00:47:31 · answer #4 · answered by yan 2 · 0⤊ 0⤋