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An object of mass (m=4.0kg) , starting from rest, slides down an inclined plane of length (l=3.0m) . ?
The plane is inclined by an angle of \(\theta=30^\circ\) to the ground. The coefficient of kinetic friction (mu_{k,1}=0.2) . At the bottom of the plane, the mass slides along a rough surface with a coefficient of kinetic friction (mu_{k,2}=0.3) until it comes to rest.
How do I find how far from the bottom of the inclined plane does the object slide along the rough surface? My answer is 250.78m. Is it correct?
1 Answer
- oubaasLv 73 weeks ago
Energy at the bottom E = m*g*L(sin 30-cos30*0.2) = 4*9.806*3*(0.5-0.866*0.2) = 38.46 joule
E = m*g*0.3*d
d = 38.46/(4*9.806*0.3) = 3.27 m
