Complex Kinematics and Dynamics

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SUMMARY

The discussion revolves around a physics problem involving two Teflon pucks, each weighing 5 kg, on a Teflon table. Puck A is initially at rest and is pushed with a force of 20 N at a 30° angle after a delay of 1.5 seconds, while Puck B travels towards it at 0.5 m/s. The calculations involve determining the net force, acceleration, and distance traveled by Puck A before it collides with Puck B, factoring in friction with a coefficient of 0.04. The final calculations yield a distance of 20.41 m for Puck A, which raises concerns about the validity of the result.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of kinematics and dynamics
  • Familiarity with forces, friction, and acceleration calculations
  • Ability to solve quadratic equations
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  • Review Newton's second law of motion and its application in dynamics
  • Study the effects of friction in motion problems, specifically with Teflon surfaces
  • Learn to apply kinematic equations in collision scenarios
  • Practice solving quadratic equations in the context of physics problems
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Vraj Patel
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Homework Statement


Two pucks (5 kg each) made of Teflon are on a long table, also made of Teflon. Puck A is sitting at
rest on the left end of the table. Puck B is 15 m away at the right hand end of the table, and is
travelling toward Puck A with an initial speed of 0.5 m/s. A person on the left waits 1.5 seconds and
then pushes Puck A forward from rest with a force of 20 N applied at an angle of 30° to the horizontal.
If that force is maintained at a steady rate for the entire question, how far will puck A travel before it
strikes puck B? (The coefficient of friction between the Teflon and Teflon is 0.04.

2. The attempt at a solution:

Puck A
Fax = 20NSin30
Fax = 10N

Fay = 20NCos30
Fay = 17.32N

Fg = mg
Fg = (5kg)(9.8m/s^2)
Fg = 49N


Fn = Fay + Fg
Fn =17.32N + 49N
Fn = 66.32

Ff = uFn
Ff = (0.04)(66.32N)
Ff = 2.65

Fnet = Fax - Ff
Fnet = 10N - 2.65
Fnet = 7.35N

a = Fnet/m
a = 7.35N/5kg
a = 1.47m/s^2

d = v1t + 1/2at^2
15-x = 1/2(1.47)(y-1.5)^24
x = -0.74y^2 + 2.22y +13.33

Puck B
Fg = mg
Fg = (5kg)(9.8m/s^2)
Fg = 49N

Ff = uFn
Ff = (0.04)(49N)
Ff = 1.96N


Since there is no applied force Fnet = -Ff

a = Fnet/m
a = -1.96/5kg
a = -0.39m/s^2

d = v1t + 1/2at^2
x = (0.5)(y) + 1/2(-0.39)(y)^2
-0.74y^2 + 2.22y +13.33 = 0.5 - 0.195y^2
0 = 0.545y^2 - 1.72y-13.33


I then put it in the quadratic formula and got:
y = 6.77s and y = -3.61s(omited)


Puck A Distance
d = v1t + 1/2at^2
d = (0m/s)(6.77s-1.5s) + 1/2(1.47m/s^2)(6.77-1.5s)^2
d = 20.41m(This answer doesn't make sense)

20181005_162939.jpg
20181005_162946.jpg


my-drive

my-drive
 

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Your "attempt at solution" is unreadable. Please type it out.
 
Your images are on your private google drive, so the links are inaccessible to other members. Either type out your solutions as suggested, or at least use the UPLOAD facility to make the images local to the PF server.
 
Doc Al said:
Your "attempt at solution" is unreadable. Please type it out.
I uploaded the photos from a local source, so you should be able to see my work. Thank you for taking the time to help me.
 
gneill said:
Your images are on your private google drive, so the links are inaccessible to other members. Either type out your solutions as suggested, or at least use the UPLOAD facility to make the images local to the PF server.
I uploaded the photos from a local source, so you should be able to see my work. Thank you for taking the time to help me.
 
Vraj Patel said:
I uploaded the photos from a local source, so you should be able to see my work. Thank you for taking the time to help me.
Your work in the images is basically unreadable. You'll have to type in your work.
 
gneill said:
Your work in the images is basically unreadable. You'll have to type in your work.
ok
 
gneill said:
Your work in the images is basically unreadable. You'll have to type in your work.
I typed out all the steps I used.
 

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