Posted by Breezerlea on .
A 10.3 kg block of ice slides without friction down a long track. The start of the track is 4.2 m higher than the end of the track, and the path traveled is 8.0 m. Find the speed of the block when it reaches the end of the track.
sinA = Y/r = 4.2 / 8 = 0.525,
A = 31.67deg.
Wb=10.3kg * 9.8N/kg = 100.9N @ 31.7deg = Weight of block.
Fp = 100.9sin31.7 = 53N = Force parallel to plane.
Ff = 0 = Force due to friction.
Fn=Fp - Ff = 53 - 0 = 53N = Net force.
a = Fn / m = 53 / 10.3 = 5.15m/s^2.
Vf^2 = Vo^2 + 2ad,
Vf^2 = 0 + 2 * 5.15 * 8 = 82.3,
Vf = 9.1m/s.