Asked by Daniel
A block of mass 2.00kg starts at rest at the top of a ramp that is inclined delta= 40.0 degrees with the horizontal. The block slides down the ramp a distance of 3.60 meters where it contacts a relaxed spring with K=300N/m. If the coefficient of kinetic friction between the block and the ramp is 0.300, find:
a) the speed of the block when it contacts the spring.
b) the distance the block compresses the spring
c) the distance between the block's initial position and the highest point it reaches after the spring sends it back up the incline
a) the speed of the block when it contacts the spring.
b) the distance the block compresses the spring
c) the distance between the block's initial position and the highest point it reaches after the spring sends it back up the incline
Answers
Answered by
Henry
Wb = mg = 2kg * 9.8N/kg = 19.6N. = Weight of block.
Fb = (19.6N,40deg.).
Fp = 19.6sin40 = 12.60N. = Force parallel to the ramp.
Fv = 19.6cos40 = 15.00N. = Force perpendicular to ramp.
Ff = u*Fv = 0.30 * 15 = 4.5N.
a. Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*3.6sin40 = 45.36,
Vf = 6.73m/s.
b. Fn = Fp - Ff = 12.6 - 4.5 = 8.1N. = Net force.
d = Fn / K = 8.1N / 300N/m = 0.027m.
Fb = (19.6N,40deg.).
Fp = 19.6sin40 = 12.60N. = Force parallel to the ramp.
Fv = 19.6cos40 = 15.00N. = Force perpendicular to ramp.
Ff = u*Fv = 0.30 * 15 = 4.5N.
a. Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*3.6sin40 = 45.36,
Vf = 6.73m/s.
b. Fn = Fp - Ff = 12.6 - 4.5 = 8.1N. = Net force.
d = Fn / K = 8.1N / 300N/m = 0.027m.
Answered by
Anonymous
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