Asked by Rachael
A block is released on an inclined plane from a height of 1m. There is friction (static and kinetic =.4) and the mass of the block is 5 kg, the mass of the pulley is 1 kg and it has a radius of 5 cm. The slope of the incline is 30 degrees.
Find the acceleration of the block as it slides down the plane.
Find the velocity of the block at the bottom of the plane.
Find the acceleration of the block as it slides down the plane.
Find the velocity of the block at the bottom of the plane.
Answers
Answered by
Henry
Wb = mg = 5kg * 9.8N/kg = 49N.
Fb = 49N @ 30 eg. = Force of block.
Fp = 49sin30 = 24.5N. = Force parallel to incline.
Fv = 49cos30 = 42.4N. = Force perpendicular to incline.
Fk = u*Fv = 0.4 * 42.4 = 17N. = Force of kinetic friction.
1. a = (Fp-Fk) / (m1+m2),
a = (24.5-17) / (5+1) = 1.25m/s^2.
2. Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*1 = 19.6,
Ff = 4.43m/s.
Fb = 49N @ 30 eg. = Force of block.
Fp = 49sin30 = 24.5N. = Force parallel to incline.
Fv = 49cos30 = 42.4N. = Force perpendicular to incline.
Fk = u*Fv = 0.4 * 42.4 = 17N. = Force of kinetic friction.
1. a = (Fp-Fk) / (m1+m2),
a = (24.5-17) / (5+1) = 1.25m/s^2.
2. Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*1 = 19.6,
Ff = 4.43m/s.
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