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

Step-by-step 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

Answer 2

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.

Step-by-step 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 =
`(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 * (-0.9)`

`F_(r)` = 7.614 N

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

`F_(r) = \mu * N`

`\mu = (F_(r))/(N)`

=
`(7.614)/(108)`

= 0.0705

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