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(Mon) A 3.00 kg block starts from rest at the top of a 30.0 º incline and accelerates uniformly down the incline moving 2.00 m in 1.50 s. What is the magnitude of the acceleration of the block? (5 min / 5 pts).

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  1. (Mon) A 3.00 kg block starts from rest at the top of a 30.0º incline and accelerates uniformly down the incline moving 2.00 m in 1.50 s. What is the magnitude of the acceleration of the block? (5 min / 5 pts)

  2. (Mon) A 3.00 kg block starts from rest at the top of a 30.0º incline and accelerates uniformly down the incline moving 2.00 m in 1.50 s. What is the magnitude of the acceleration of the block? (5 min/5 pts) 2.0 m N/A 0 m/s N/A N/A ????? N/A Δx = vit + ½ at² 1.5 s N/A 2.0 m = 0 + ½ a(1.5s)² 2.0 m / (0.5)(2.25 s²) = a 1.78 m/s² = a

  3. (Tue) A 3.00 kg block starts from rest at the top of a 30.0º incline and accelerates uniformly down the incline moving 2.00 m in 1.50 s. If the acceleration down the block is 1.78 m/s², what is the magnitude for the net force for the block down the ramp? (5 min / 5 pts)

  4. (Tue) A 3.00 kg block starts from rest at the top of a 30.0º incline and accelerates uniformly down the incline moving 2.00 m in 1.50 s. If the acceleration down the block is 1.78 m/s², what is the magnitude of the net force for the block down the ramp? (5 min/5 pts) F = ma F = (3.00 kg) x (1.78 m/s²) F = 5.34 kgm/s² F = 5.34 N

  5. (Wed) A 3.00 kg block starts from rest at the top of a 30.0º incline and accelerates uniformly down the incline moving 2.00 m in 1.50 s. If the net force down the ramp is 5.34 N, what is the force due to friction acting on the block? (5 min / 5 pts)

  6. (Wed) A 3.00 kg block starts from rest at the top of a 30.0º incline and accelerates uniformly down the incline moving 2.00 m in 1.50 s. If the net force down the ramp is 5.34 N, what is the force due to friction acting on the block? (5 min/5 pts) 14.71 N 0 N ???? In the ‘x’ direction….. 5.34 N Fnet = Fg + Fn + Ff Fg,x = magsin(30) 5.34 N = 14.71 N + 0 + Ff Fg,x = (3.00 kg)(9.81 m/s²)(0.5) -9.37 N = Ff Fg,x = 14.71 N

  7. (Thu) A 3.00 kg block starts from rest at the top of a 30.0º incline and accelerates uniformly down the incline moving 2.00 m in 1.50 s. If the force due to friction is -9.37 N and the net force down the ramp is 5.34 N, what is the coefficient of friction between the block and the ramp? (8 min / 5 pts)

  8. (Thu) A 3.00 kg block starts from rest at the top of a 30.0º incline and accelerates uniformly down the incline moving 2.00 m in 1.50 s. If the force due to friction is -9.37 N and the net force down the ramp is 5.34 N, what is the coefficient of friction between the block and the ramp? (8 min/5 pts) In the ‘y’ direction Fnet = Fg + Fn + Ff -25.49 N 0.0 N 0 N = -25.49 N + Fn + 0 N 25.49 N 25.49 N = Fn -9.37 N 0.0 N 5.34 N 0.0 N Ff = (Fn)(μs) (μs) = 0.37 Fg,y = magcos(30) Ff/(Fn) = (μs) Fg,y = (3.00kg)(9.81 m/s²)(0.86) (μs) = 9.37 N / 25.49 N Fg,y = 25.49 N

  9. (Fri) A 3.00 kg block starts from rest at the top of a 30.0º incline and accelerates uniformly at a rate of 1.78 m/s² down the incline moving 2.00 m in 1.50 s. What is the magnitude of the velocity for the block at the bottom after traveling the 2.00 m? (5 min / 5 pts)

  10. (Fri) A 3.00 kg block starts from rest at the top of a 30.0º incline and accelerates uniformly at a rate of 1.78 m/s²down the incline moving 2.00 m in 1.50 s. What is the magnitude of the velocity for the block at the bottom after traveling the 2.00 m? (5 min/5 pts) N/A 2.0 m 0 m/s N/A N/A ????? vf² = vi² + 2aΔx N/A 1.78 m/s² vf² = 0² + 2(1.78 m/s²)(2.0 s) N/A 1.5 s vf² = 7.12 m²/s² vf = 2.67 m/s

  11. End of Week Procedures • Add up all the points you got this week • Put the total number of points out of 25 points you earnedat the top of your page (no total, lose 5 points) • List any dates you were absent or tardy (and why) • Make sure your name and period # is on the top of the page (no name, no credit) • Turn in your papers

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