1 / 11

Physics Problem Solving: Block Acceleration and Forces

Solve a physics problem involving a 3.00 kg block accelerating down a 30.0º incline, calculating acceleration, net force, friction force, coefficient of friction, and final velocity.

ecausey
Download Presentation

Physics Problem Solving: Block Acceleration and Forces

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related