Question

An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane is 10 m long and the mass starts from rest, what will be its speed at the bottom of the plane?

Answer 1

`v \approx 9.312\,(m)/(s)`

Step-by-step explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

`K_(A) + U_(g,A) = K_(B) + U_(g,B) + W_(loss)`

`K_(B) = K_(A) + U_(g,A)-U_(g,B) - W_(loss)`

`(1)/(2)\cdot m \cdot v^(2) = m\cdot g \cdot s\cdot \sin \theta - \mu_(k)\cdot m \cdot g \cdot s \cos \theta`

`(1)/(2)\cdot v^(2) = g\cdot s \cdot (\sin \theta - \mu_(k)\cdot \cos \theta)`

`v = \sqrt{2\cdot g \cdot s \cdot (\sin \theta - \mu_(k)\cdot \cos \theta)}`

`v = \sqrt{2\cdot (9.807\,(m)/(s^(2)) )\cdot (10\,m)\cdot (\sin 37^(\textdegree) - 0.2\cdot \cos 37^(\textdegree))}`

`v \approx 9.312\,(m)/(s)`