Answered

Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

A box slides down a 28.0 degree ramp with an acceleration of 1.25 m/s2. Determine the coefficient of kinetic friction between

Sagot :

okpalawalter8

Answer:

[tex]\mu=0.39[/tex]

Explanation:

From the question we are told that:

Angle [tex]\theta=28[/tex]

Acceleration [tex]a=1.25m/s^2[/tex]

Generally the equation for Frictional force  is mathematically given by

[tex]F=\muN[/tex]

Where

[tex]N=mgcos \theta[/tex]

[tex]N=mgcos 28[/tex]

Since

Friction force is acting against move of box

Therefore

[tex]mgsin(28) - 1.25m = \mu mgcos(28)[/tex]

[tex]\mu=\frac{gsin(28) - 1.25}{gcos(28)}[/tex]

[tex]\mu=0.39[/tex]

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.