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(Mon) A kg block starts from rest at the top of a 30

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1 (Mon) A 3. 00 kg block starts from rest at the top of a 30
(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
(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) x Dir y Dir Δx vi vf a t 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) Veteran’s Day – No School

4 (Wed) A 3. 00 kg block starts from rest at the top of a 30
(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 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)

5 (Wed) A 3. 00 kg block starts from rest at the top of a 30
(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 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 = kgm/s² F = N

6 (Thu) A 3. 00 kg block starts from rest at the top of a 30
(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 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)

7 (Thu) A 3. 00 kg block starts from rest at the top of a 30
(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 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 x Dir y Dir Fg Fn Ff Fnet Fnet = Fg + Fn + Ff N 0.0 N 0 N = 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 / N Fg,y = N

8 (Fri) A 3. 00 kg block starts from rest at the top of a 30
(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)

9 (Fri) A 3. 00 kg block starts from rest at the top of a 30
(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) x Dir y Dir Δx vi vf a t 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

10 End of Week Procedures:
Add up all the points you got this week Put the total number of points at the top of your page (out of 20 points) List any dates you were absent (and why) Make sure your name and period is on the top of the paper Turn in your papers


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