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

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Physics Gia Bảo 3 years 2021-07-14T07:00:35+00:00 2021-07-14T07:00:35+00:00 1 Answers 8 views 0

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minhkhang · Answer 1 · 2021-07-14T07:01:43+00:00

minhkhang

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2021-07-14T07:01:43+00:00 July 14, 2021 at 7:01 am

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Answer:

$\mu=0.39$

Explanation:

From the question we are told that:

Angle $\theta=28$

Acceleration $a=1.25m/s^2$

Generally the equation for Frictional force is mathematically given by

$F=\muN$

Where

$N=mgcos \theta$

$N=mgcos 28$

Since

Friction force is acting against move of box

Therefore

$mgsin(28) - 1.25m = \mu mgcos(28)$

$\mu=\frac{gsin(28) - 1.25}{gcos(28)}$

$\mu=0.39$