Question

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward.

Answer 1

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    $F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$

             W = 83 N

         m = $\frac{83}{9.81}$

             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + $F_{2}$

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       $F_{1} - F_{r}$ = ma

        $20 - F_{r} = 8.46 \times (-0.9)$

             $F_{r}$ = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

             $F_{r} = \mu \times N$

          $\mu = \frac{F_{r}}{N}$

                    = $\frac{7.614}{108}$

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

Answer 2

Answer:

Explanation:

Weight of box, mg = 83 N

acceleration, a = - 0.9 m/s²

Horizontal component of force, Fx = 20 N

vertical component of force, Fy = - 25 N

Let f be the force of friction and μ is the coefficient of friction.

Let N be the normal reaction, by the equilibrium of forces along Y axis

N - fy = mg

N = mg + Fy

N = 83 + 25 = 108 N

Use Newton's second law

Fx - f = ma

20 - f = - 83 x 0.9 / 9.8

f = 27.6 N

f = μ N

27.6 = μ x 108

μ = 0.255