Question

A horizontal 52 Newton force is needed to slide a 50-kg box across a flat surface of a constant velocity of 3.5 m s what is the coefficient of kinetic friction between the box and the floor

Answer 1

Answer:

μk = 0.106

Explanation:

When the box is moving across the flat surface with a constant velocity, the force applied on it must be equal to the kinetic frictional force applied on the object:

F = μk*W = μk*mg

Where,

F = Force applied on box = 52 N

m = mass of box = 50 kg

g = 9.8 m/s²

μk = coefficient of kinetic friction between box and floor = ?

Therefore,

52 N  = μk(50 kg)(9.8 m/s²)

μk = (52 N)/(490 N)

μk = 0.106

Answer 2

The coefficient of kinetic friction between the box and the floor will be 0.106.

What is kinetic friction?

A force that acts among sliding parts is referred to as kinetic friction. A body moving on the surface is subjected to a force that opposes its progressive motion

The size of the force will be determined by the kinetic friction coefficient between the two materials.

The given data in the problem is;

μ is the coefficient of kinetic friction=?

m is the mass = 50 kg

g is the acceleration due to gravity= 9.81 m/s²

v is the speed =3.5 m/sec

Force(F)= 52 Newton

The formula for kinetic friction force is;

$\rm F= \mu_k R \\\\ R=mg \\\\ F= \mu_k mg \\\\\ 52 = \mu_k \times 50 \times 9.81 \\\\ \mu_k = 0.106$

Hence,the coefficient of kinetic friction between the box and the floor will be 0.106.

To learn more about the kinetic friction refer to;

https://brainly.com/question/13754413

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