Question

A 23-N force is applied to a 7-kg object to move it with a constant velocity of 6.1 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

Given:

The applied force is F = 23 N

The mass of the object is m = 7 kg

The object moves with a constant velocity of v = 6.1 m/s

The acceleration due to gravity is g = 10 m/s^2

Required: The coefficient of friction between the object and the surface.

Step-by-step explanation:

The object moves at a constant velocity, so the acceleration will be zero.

The equation of motion can be written as

`\begin{gathered} F-f\text{ = ma} \\ F-f\text{ =m}*0 \\ F-f=0 \\ F=f \end{gathered}`

So, the applied force will be equal to the frictional force.

The coefficient of friction can be calculated as

`\begin{gathered} F\text{ = }\mu mg \\ \mu=(F)/(mg) \end{gathered}`

On substituting the values, the coefficient of friction will be

`\begin{gathered} \mu=(23)/(7*10) \\ =\text{ 0.33} \end{gathered}`

Thus, the coefficient of friction is 0.33

Final Answer: The coefficient of friction between the object and the surface is 0.33.