A block starts from rest at a height of 8.6m on a fixed inclined plane. The acce

Question

A block starts from rest at a height of 8.6m on a fixed inclined plane. The acceleration of gravity is 9.8m/s^2.
Part 1. What is the speed of the block at the bottom of the ramp? Answer in units of m/s
Part 2. If the block containues to slide on the ground with the same coefficient of friction, how far will the block slide on the ground untill coming to rest? Answer in units of m.
Block wieght, 5.5kg
µ=0.28 Degree- 23 A block starts from rest at a height of 8.6m on a fixed inclined plane. The acceleration of gravity is 9.8m/s^2.
Part 1. What is the speed of the block at the bottom of the ramp? Answer in units of m/s
Part 2. If the block containues to slide on the ground with the same coefficient of friction, how far will the block slide on the ground untill coming to rest? Answer in units of m.
Block wieght, 5.5kg
µ=0.28 Degree- 23

Explanation / Answer

mass m = 5.5 kg coeffcient of friciton u = 0.28 angle theta = 23 degrees height of the ramp h = 8.6 m Initial velocity of the block U = 0 accleration a =? net force F = mg sin(theta) - umg cos(theta) where umg cos(theta) is frictional force accleration a = F / m = g sin 23 - 0.28*g cos23 = 3.829 - 2.525 = 1.304 m / s^ 2 length of the ramp L = h / sin23 = 22 m the speed of the block at the bottom of the ramp v = sqrt[ U^ 2+2aL] = 7.576 m / s on ground: ---------- initial speed u ' = 7.576 m / s final speed v' = 0 accleration a ' = -ug = -2.744 m / s^ 2 distance travelled before stop S = [ v'^ 2-u'^ 2] / 2a' = 10.45 m

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