Question

A 33-N force is applied to a 7-kg object to move it with a constant velocity of 6.3 m/s across a level surface. The coefficient of friction between the object and the surface is approximately ____. Round your answer to the hundredths place.(Use the approximation g ≈ 10 m/s2.)

Answer 1

The coefficient of friction between the object and the surface is approximately 0.47.

Step-by-step explanation:

The force of friction can be determined by using the equation:

Ffriction = μN

where μ is the coefficient of friction and N is the normal force.

In this case, the normal force is equal to the weight of the object:

N = mg

where m is the mass of the object and g is the acceleration due to gravity (approximately 10 m/s2).

Therefore, the force of friction can be calculated as:

Ffriction = μmg

Since the object is moving with a constant velocity, the force of friction must be equal in magnitude and opposite in direction to the applied force:

Ffriction = Fapplied

Using the given values of Fapplied (33 N) and the mass of the object (7 kg), we can solve for the coefficient of friction:

μmg = Fapplied

μ = Fapplied/mg

Substituting the values into the equation:

μ = 33 N / (7 kg * 10 m/s2)

Therefore, the coefficient of friction is approximately 0.47.

Answer 2

Given:

The applied force is,

`F=33\text{ N}`

The mass of the object is,

`m=7\text{ kg}`

The constant velocity of the object is,

`v=6.3\text{ m/s}`

To find:

The coefficient of friction between the object and the surface

Explanation:

The object is moving with a constant velocity which means the object is under equilibrium. So, the applied force is equal to the frictional force on the object.

If the coefficient of friction between the object and the surface is

`\mu`

we can write the frictional force as,

`f=\mu mg`

For equilibrium condition,

`\begin{gathered} \mu mg=F \\ \mu=(F)/(mg) \\ \mu=(33)/(7*10) \\ \mu=0.47 \end{gathered}`

Hence, the coefficient of friction between the object and the surface is 0.47.