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Savatey [412]
3 years ago
10

Once the crate in sample problem d is in motion, a horizontal force of 53 n keeps the crate moving with constant velocity find t

he coefficient of kinetic friction between the crate and the floor

Physics
2 answers:
IgorC [24]3 years ago
4 0
Thank you for posting your question here at brainly. Below is the solution. I hope it helps. 

m=24 kg, Ff=53 N, g=9.8 m/s^2,
use the formula Fg<span> = mg,   F</span>g=24*9.8,
solve for Fg, Fg=235.2 N,
then use the formula Ff<span> = ต</span>kFn, 53=235.2ตk,
solve for ตk, ตk<span>=.23</span>
mixer [17]3 years ago
4 0

<u>The coefficient of kinetic friction is 0.225. </u>

Further explanation:

Friction force is defined as the opposing force acting in the opposite direction of the motion of the body.

Given:

The magnitude of kinetic frictional force is 53\,{\text{N}}.

The mass of the body is 24\,{\text{Kg}}.

Concept used:

A force is applied on the crate to displace it from one place to another. There is opposing force acting in the opposite direction to the motion of the crate known as friction force.

The expression for the kinetic friction is given as.

{F_k} = {\mu _k}N                                                   …… (1)

Here, {F_k} is the kinetic friction force, {\mu _k} is the coefficient of kinetic friction and N is the normal force.

The normal force acting on the body is balanced by its weight.

The expression for the normal force is given as.

N=mg

Here, m is the mass of the body and g is the acceleration due to gravity.

Substitute mg for N in equation (1).

{F_k}={\mu _k}\left({mg}\right)

 

Rearrange the above expression for the coefficient of kinetic friction.

\fbox{\begin\\{\mu _k}=\dfrac{{{F_k}}}{{\left( {mg}\right)}}\end{minispaace}}                                             …… (2)

Substitute 53\,{\text{N}} for {F_k}, 24\,{\text{Kg}} for m and 9.8\,{\text{m/}}{{\text{s}}^{\text{2}}} for g in equation (2).

\begin{aligned}{\mu _k}&=\frac{{53\,{\text{N}}}}{{\left( {\left( {24\,{\text{Kg}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)} \right)}}\\&=0.225\\\end{aligned}

 

Thus, the coefficient of kinetic friction is \fbox{0.225}.

Learn more:

1.  Motion under friction brainly.com/question/7031524.

2.  Conservation of momentum brainly.com/question/9484203.

3. Net force of body brainly.com/question/6125929.

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Acceleration, force, acceleration due to gravity, friction, normal, weight, mass, motion, sliding, sled, hill, inclined, plane, coefficient of friction, angle of inclination, 0.225, 0.23, 0.2254.  

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