Question

A 5.00 kg box is being pulled at constant speed up a 30o incline by a force of 45.0 N parallel to the incline. Determine the coefficient of kinetic friction between the box and the plane.

Answer 1

Answer:

the box will 200g

Explanation:

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Answer 2

Given: Mass of block = 5kg

           Inclination of slope = 30°

          Force being applied in upward direction = 45 N

To Find: Coefficient of kinetic friction between box and plane

Solution:

• When we make the FBD (free body diagram) for this situation we find that :

The force mg has 2 components:

1. mgsinФ which, along with friction f balances out F=45 N.(x-axis)
2. mgcosФ which balances out Normal. (y-axis)

Using the components of mg we will find the coefficient of kinetic friction.

                             mgsinФ + f = F

                          ⇒ 5 × 10×$\frac{1}{2}$ + f = 45                 (1)  (sin 30 = $\frac{1}{2}$)

Now, friction can be found using the formula:

                       f = coefficient of friction × normal

                  ⇒ f = μ × mgcosФ                   (normal is balanced by mgcosФ)

Substituting this value of friction in (1)

                     ⇒   25 + μ × 25 × √3 = 45               (cos 30 = $\frac{\sqrt{3} }{2}$)

                     ⇒   25(1 + √3μ) = 45

                     ⇒  1 + √3μ = $\frac{9}{5}$                                  (dividing 45 by 25)

                     ⇒   1 + √3μ = 1.8

                     ⇒  √3μ = 0.8

                     ⇒ μ = 0.46

Hence, coefficient of kinetic friction between box and plane is 0.46.