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
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?
1 answer
0 answers