Corkscrew
This presentation is the property of its rightful owner.
Sponsored Links
1 / 11

CORKSCREW PowerPoint PPT Presentation


  • 58 Views
  • Uploaded on
  • Presentation posted in: General

CORKSCREW. QUESTION: The roller coaster is at the top of the hill 20m high and moving at a speed of 10m/sec. a ) (assume no friction) How fast will the coaster be moving at the bottom of the hill before going into the loop?.

Download Presentation

CORKSCREW

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Corkscrew

CORKSCREW


Question the roller coaster is at the top of the hill 20m high and moving at a speed of 10m sec

QUESTION: The roller coaster is at the top of the hill 20m high and moving at a speed of 10m/sec.


Corkscrew

a) (assume no friction) How fast will the coaster be moving at the bottom of the hill before going into the loop?


Corkscrew

b) (assume no friction) How fast will the roller coaster be moving at the top of a loop which is 15m high?


Corkscrew

The density of air is 1.75 kg/m3, the area of the coaster is 4m2, and the drag coefficient for the coaster is 0.8.


Corkscrew

c) (in a perfect physics world) If the wind starts up after going out of the loop; a drag force is created to stop the coaster. Find the deceleration of the coaster if the coaster is running on a flat track.


Find the deceleration of the coaster starting here

Find the deceleration of the coaster starting here!


Corkscrew

ANSWERS


Corkscrew

A

Since gravitational potential energy is 0 at the bottom; PE and mass cancels out.

½v2 = ½v2

250 = ½v2

2 x 250 = v2

√500 = v

22.36 = v


Corkscrew

B

PE + KE = PE + KE

Mgh + 1/2mv2 = mgh + 1/2mv2

(9.8)(20) + ½(10)2 = (9.8)(15) + ½ v2

196 + 50 = 147 + ½ v2

2(246-147) = v2

√198 = v

14.07 = v


C fd ma pv 2 c d a ma 1 75 22 36 2 0 8 4 400a 1399 91488 400a 1399 91488 400 a 3 50 m s 2 a

CFd = ma½pv2CdA = ma ½(1.75)(22.36)2(0.8)(4) = 400a1399.91488 = 400a1399.91488/400 = a3.50 m/s2 = a


  • Login