A block weights 150g and it’s attached to a 0.6m long string. The block starts from rest and begins to spin in a circle accelerating at a constant rate of 0.25 rad/s^2 on a horizontal surfaces if the string has a max tension of 5N. How long does it take for the string to break? How many revolutions did the block complete? Please Use the equations provided.

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter2: One Dimensional Motion
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Problem 46PQ: In Example 2.6, we considered a simple model for a rocket launched from the surface of the Earth. A...
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A block weights 150g and it’s attached to a 0.6m long string. The block starts from rest and begins to spin in a circle accelerating at a constant rate of 0.25 rad/s^2 on a horizontal surfaces if the string has a max tension of 5N. How long does it take for the string to break? How many revolutions did the block complete? Please Use the equations provided.
Equations:
Vector components, standard Cartesian plane:
Vx= V cos 8
Vy = V sin 0
Constant linear acceleration:
V = Vo + at
Projectile motion:
X = Xo + Voxt
Rotational motion:
0 = s/r
@= @o + at
Frictional force:
w = v/r
Newton's Second Law of motion:
Centripetal force:
Use g
d = do + Vot + ½ at²
Gravitational force, or weight:
=
Vx² + Vy² = V²
y = yo + Voy - 1/2 gt²
360⁰ = 2n radians
a = aT/r
000+ wot + ½ at²
tan 8 = Vy/Vx
v² = vo² + 2a(d-do)
Vy = Voy -gt
ac = v²/r = rw²
w² = wo² + 2a(0 - 0)
ΣF = ma
Fg = mg
Ff = µFn
Fc = mv²/r
10.0 m/s² for the acceleration due to gravity near the surface of the earth.
Transcribed Image Text:Equations: Vector components, standard Cartesian plane: Vx= V cos 8 Vy = V sin 0 Constant linear acceleration: V = Vo + at Projectile motion: X = Xo + Voxt Rotational motion: 0 = s/r @= @o + at Frictional force: w = v/r Newton's Second Law of motion: Centripetal force: Use g d = do + Vot + ½ at² Gravitational force, or weight: = Vx² + Vy² = V² y = yo + Voy - 1/2 gt² 360⁰ = 2n radians a = aT/r 000+ wot + ½ at² tan 8 = Vy/Vx v² = vo² + 2a(d-do) Vy = Voy -gt ac = v²/r = rw² w² = wo² + 2a(0 - 0) ΣF = ma Fg = mg Ff = µFn Fc = mv²/r 10.0 m/s² for the acceleration due to gravity near the surface of the earth.
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