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DENIUS [597]
3 years ago
11

A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep

the box moving at a constant velocity. What is the coefficient of kinetic friction between the box and the floor?
0.23

0.20

0.18

0.17
Physics
1 answer:
vitfil [10]3 years ago
7 0

Answer:

0.20

Explanation:

The box is moving at constant velocity, which means that its acceleration is zero; so, the net force acting on the box is zero as well.

There are two forces acting in the horizontal direction:

- The pushing force: F = 99 N, forward

- The frictional force: F_f=\mu mg, backward, with

\mu coefficient of kinetic friction

m = 50 kg mass of the box

g = 9.8 m/s^2 gravitational acceleration

The net force must be zero, so we have

F-F_f = 0

which we can solve to find the coefficient of kinetic friction:

F-\mu mg=0\\\mu = \frac{F}{mg}=\frac{99 N}{(50 kg)(9.8 m/s^2)}=0.20

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