Question
A 6 kg black is placed at the very top of a ramp that is inclined at 31 degrees and the block is released to slide down the ramp. As the block slides friction exerts a constant resistive force of 17N directed back up the ramp
What is the block's speed when it arrives at the bottom of the 3.5 meter long ramp?
What is the block's speed when it arrives at the bottom of the 3.5 meter long ramp?
Answers
the gravitational potential energy at the top of the ramp becomes kinetic energy at the bottom of the ramp (minus the work done by friction)
m g h - f d = 1/2 m v^2
[6 * 9.8 * 3.5 sin(31º)] - (17 * 35.) = 1/2 * 6 * v^2
m g h - f d = 1/2 m v^2
[6 * 9.8 * 3.5 sin(31º)] - (17 * 35.) = 1/2 * 6 * v^2