Question

What is the mu of an object that slides down an inclined plane (Angle 41 degrees) and has a speed of 2m/s in 4 seconds?

Answer 1

\[\mu =\tan \theta\], the velocity is constant?

Answer 2

http://www.physicsclassroom.com/class/vectors/u3l3e.cfm

Answer 3

We know that
\[F_{friction}=\mu\times F_{normal}\]Where\[F_{normal}=m\times g\times\cos\theta\]\[F_{tangential}=m\times g\times \sin\theta\]
Also along the incline\[F_{eff}=F_{tangential}-F_{friction}\]\[F_{eff}=m\times a_{eff}\]
By laws of motion\[v_{final}=u+a_{eff}\times t\]Where\[t=4\ s\]\[u=0\ m/s\]\[v_{final}=2\ m/s\]Thus\[a_{eff}=2/4 = 0.5\ m/s^2\]
For the object\[m\times0.5=m\times g\times\sin\theta-\mu\times m\times g\times\cos\theta\]Where\[\theta=41\ deg\]\[\sin\theta=0.6561\]\[\cos\theta=0.7547\]
Solving for mu\[0.5=g\times(0.6561 - \mu\times0.7547)\]\[\mu\times0.7547=0.6561-0.05097\]\[\mu=0.8018\]