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A 8.00kg block of ice, released from rest at the top of a 1.50m long frictionless ramp, slides downhill, reaching a speed of 2.50 m/s at the bottom. What would be the speed of the cie at the bttom if the motion were opposed by a constant friction force of 10.0N parallel to the surface of the ramp?
The unknown angle of the ramp is 12.3 degrees
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Added Wed, 22 Jul '15
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Work Done = Change in Kinetic Energy

Loss in Gravitational Energy = Work Done Against Friction + Gain in Kinetic Energy

mgh = Frictional Force*Length of Ramp + 0.5*m*v^2
As we know that Sin(theta) = Perpendicular/Hypotenuse

Here Perpendicular = height = h

and

Hypotenuse = Length of Ramp = 1.5 m.

Therefore h = 1.5sin(12.3)

Therefore

mgh = Frictional Force*Length of Ramp + 0.5*m*v^2

8*9.8*1.5*Sin(12.3 degree) = 10*1.5 + 0.5*8*v^2

v = 1.585 m/sec
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Added Wed, 22 Jul '15
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