Hint: This problem requires a train of logic. (1) Analyze force diagram, (2) use Newton's Laws, and (3) solve the equations of motion. A block starts from rest at a height of 3.9 m on a fixed inclined plane. The acceleration of gravity is 9.8 m/s 2. 3.8 kg µ = 0.12 31◦ What is the speed of the block at the bottom of the ramp? Answer in units of m/s.

Question

Olive Gonzales

Physics

21 December, 17:00

Hint: This problem requires a train of logic. (1) Analyze force diagram, (2) use Newton's Laws, and (3) solve the equations of motion. A block starts from rest at a height of 3.9 m on a fixed inclined plane. The acceleration of gravity is 9.8 m/s 2. 3.8 kg µ = 0.12 31◦ What is the speed of the block at the bottom of the ramp? Answer in units of m/s.

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Xzavier Donaldson · Answer 1

8. 55599 m / s.

Let : h = 8. 2 m, m = 5. 1 kg, = 0. 22, = 22, and v f = final speed. The normal force to the inclined plane is N = mg cos. The sum of the forces par - allel to the inclined plane is F net = ma = mg sin - mg cos a = g sin - g cos Since v 2 f = v 2 + 2 ax = 2 ad (1) along the plane, and neglecting the dimension of the block, the distance, d, to the end of the ramp is d sin = h d = h sin (2) therefore v f = r 2 ah sin = r 2 g h (sin - cos) sin = p 2 g h (1 - cot) (3) = q 2 (9. 8 m / s 2) (8. 2 m) [1 - (0. 22) cot22 ] = 8. 55599 m / s.