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Write an algebraic expression for the coefficient of kinetic friction for a block of mass m sliding down an incline of angle theta with respect to the horizontal with acceleration a. (type the formula out with the word (theta) for the angle. Make sure you are careful with your parentheses. You may use the following variables: m, a, g, (theta) and the sin, cos, and tan function

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abidemiokin

Answer:

n = {gsin(theta) - a}/gcos(theta)

OR n = tan(theta) {at constant velocity}

Explanation:

For a block of mass m sliding down an incline at angle theta, the forces acting on the object along the plane are the moving force (down the plane) and the frictional force which is acting opposite the moving force (i.e acting up the plane).

The moving force Fm = Wsin theta where W is the weight of the block and theta is the angle of inclination.

The frictional force Ff is expressed as the product of the coefficient of kinetic friction (n) and the normal reaction force (R) acting on the body i.e Ff = nR

Since the body is accelerating down the plane, we will apply the Newton's second law to solve for the coefficient of the kinetic friction.

According to the law;

Force = mass × acceleration

F = ma... (1)

Summation of the force acting on the body along the plane will be;

Fm - Ff

= Wsin (theta) - nR

Since R =Wcos(theta), the sum of the forces becomes;

Wsin(theta) - n(Wcos(theta))

Since W = mg, the equation becomes;

mgsin(theta) - nmgcos(theta)... (2)

Substituting equation 2 into 1, we will have;

mgsin(theta) - nmgcos(theta) = ma

gsin(theta) - ngcos(theta) = a

ngcos(theta) = gsin(theta) - a

Dividing both sides by gcos(theta) to get n which is the coefficient of kinetic friction will give;

n = {gsin(theta) - a}/gcos(theta)

If the body moves with a constant velocity, acceleration 'a' will be zero and the coefficient of kinetic friction n will become;

n = {gsin(theta) - 0}/gcos(theta)

n = gsin(theta)/gcos(theta)

n = sin(theta)/cos(theta)

n = tan(theta)

snehashish65

The algebraic expression for the coefficient of kinetic friction for a block of mass m sliding down an incline plane is, [tex]n = tan \theta[/tex].

Kinetic friction:

The Force of friction comes to play, during motion of an object is known as kinetic friction.

As per the given problem,

  • The block of mass m sliding down an incline at angle [tex]\theta[/tex], the forces acting on the object along the plane are the moving force (down the plane) and the frictional force which is acting opposite the moving force (i.e acting up the plane).

  • The moving force [tex]F= W sin \theta[/tex],  where W is the weight of the block and theta is the angle of inclination.

  • The frictional force f is expressed as the product of the coefficient of kinetic friction (n) and the normal reaction force (R) acting on the body i.e f = nR.

  • Since the body is accelerating down the plane, we will apply the Newton's second law to solve for the coefficient of the kinetic friction.

        According to the law;

        F = ma............................................ (1)

  • Sum of the force acting on the body along the plane will be;

       [tex]F -f = W sin \theta -nR[/tex]  

        Since,

      [tex]R = W cos \theta[/tex]

  • Then, the sum of the forces becomes;

        [tex]W sin \theta - n(W cos \theta) = ma\\\\(mg) sin \theta - n((mg) cos \theta) = ma\\\\gsin \theta - ng cos \theta = a[/tex]

        Here, a is the linear acceleration and for the constant velocity, a = 0.      

        [tex]gsin \theta - ng cos \theta = 0\\\\ngcos \theta = g sin \theta\\\\n = tan \theta[/tex]

Thus, we can conclude that the algebraic expression for the coefficient of kinetic friction for a block of mass m sliding down an incline plane is, [tex]n = tan \theta[/tex].

Learn more about the kinetic friction here:

https://brainly.com/question/17237604

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