Asked by Breezerlea
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.
Answers
Answered by
Henry
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.
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.
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