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tankabanditka [31]
3 years ago
9

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 c

oefficient 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
Physics
2 answers:
m_a_m_a [10]3 years ago
6 0

Answer:

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

Dmitry_Shevchenko [17]3 years ago
5 0

Answer:

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

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We could take the easy way out and just say

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