Find the network done by friction on a box that moves in a complete circle of radius 1.82 m on a uniform horizontal floor. The coefficient of kinetic friction between the floor and the box is 0.25, and the box weighs 65.0 N. A) - 0 B) - 370JC) - 190 J D-1800 J

Question

Physics

Caitlin Armstrong

19 October, 02:57

Find the network done by friction on a box that moves in a complete circle of radius 1.82 m on a uniform horizontal floor. The coefficient of kinetic friction between the floor and the box is 0.25, and the box weighs 65.0 N. A) - 0 B) - 370JC) - 190 J D-1800 J

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Answers (2)

Know the Answer?

Jessica Wood · Answer 1

C) W = - 190 J

Explanation:

Notation

Wf = work done by the friction force (unknown)

Ff = force of the friction

d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m

Aydin Case · Answer 2

the network done by friction on a box that moves in a complete circle is 185.7 joules

Explanation:

Step one

Given

Radius of circle = 1.82m

Circumference of the circle = 2*pi*r

=2*3.142*1.82=11.43

Hence distance = 11.43m

Coefficient of friction u=0.25

Weight of box = 65N

We know that work = force*distance

But the limiting force = u*weight

Hence the net work done by friction

Wd=0.25*65*11.43

Wd=185.7 joules