A block sliding down an incline

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 9K views
Momentum09
Messages
69
Reaction score
0
A block starts from rest at a height of 4.7m on a fixed inclined plane tilted at 30 degrees. The coefficient of friction is 0.28. If the block continues to slide on the ground with the same coefficient of friction, how far will the block slide on the ground until coming to rest?


2. V = V0+ at



3. I found out the speed of the block at the bottom of the ramp, which is equal to 6.887937 m/s. I led Vf = 0, V0 = 6.887937, and a =2.5236 [solved from gsin(delta) - ugcos(delta). Solve for t, and plug this value into the delta x = volt + 1/2 at^2 equation. I am not sure if I'm doing the right thing.

Thank you so much!
 
Physics news on Phys.org
How did you get to that speed at the bottom of the ramp? Show your calculations please.
 
3. I found out the speed of the block at the bottom of the ramp, which is equal to 6.887937 m/s. I led Vf = 0, V0 = 6.887937, and a =2.5236 [solved from gsin(delta) - ugcos(delta). Solve for t, and plug this value into the delta x = volt + 1/2 at^2 equation. I am not sure if I'm doing the right thing.

Thank you so much!


Either calculate the force due to friction and hence the deceleration and use [tex]\ \ v^2=u^2+2as[/tex] or use the work energ concept, [tex]\ \ W_{nc}=E_f-E_i[/tex], where W_nc is the work due to friction and [tex]\ \E_f-E_i[/tex] will be same as [tex]K_f-K_i[/tex]
 
Last edited: