Question #182047

A block on a ramp slides down a 6.28 m long ramp. The ramp has an angle of inclination of 38°. If the block starts from rest and the amount of energy lost due to friction along the ramp is 21.6m J, where m is the mass of the block, use energy concepts to determine how fast the block is going at the bottom of the ramp. 


1
Expert's answer
2021-04-19T07:24:45-0400

Using energy conservation

KE at bottom = initial pE - frictional loss

12×mv2=mg×(Hsin38h)\frac{1}{2} \times m v^2 = mg\times(Hsin 38 - h)

12×mv2=m×9.8×6.28sin3821.6\frac{1}{2} \times m v^2 = m \times 9.8 \times 6.28 sin 38 - 21.6

v=2(9.8×6.28sin3821.6)v=\sqrt{2 (9.8 \times 6.28sin 38 -21.6)}

v=5.708m/sv = 5.708 m/s


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