An object of mass 4.00 kg is sitting at the top of an inclined plane of height h and angle theta with the horizontal. The object slides down the inlcined plane and right before it reaches the bottom of the plane, it has speed of 16 m/s. Neglect friction and use g = 10.0 m/s2 (second power). The height of the inclined plane is then raised to 15.0 m.What is the speed of the object right before it reaches the bottom of the inclined plane?Since friction is neglected, use conservation of potential and kinetic energy:
Energy at top of inclined plane= m*g*h
Energy at bottom of plane = 0.5*m*v^2
From conservation: v=sqrt(2*g*h)
Substituting values: v=sqrt(300) = 17.32 m/s
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