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Lecture05

# Lecture05 - PHYSICS 220 Lecture 05 Free Fall and Apparent...

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Lecture 5 Purdue University, Physics 220 1 Lecture 05 Free Fall and Apparent Weight Textbook Sections 4.1 - 4.3, 4.5-4.6 PHYSICS 220

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Lecture 5 Purdue University, Physics 220 2 Overview • Last Lecture – Net force and Newton’s second law – Relative velocity – Inertial frame of reference – Constant acceleration • This Lecture – More on constant acceleration – Gravity as a force – Free fall – Weightless
Lecture 5 Purdue University, Physics 220 3 Constant Acceleration x = x 0 + v 0 t + 1/2 at 2 v = v 0 + at v 2 = v 0 2 + 2a(x-x 0 ) Δ x = v 0 t + 1/2 at 2 Δ v = at v 2 = v 0 2 + 2a Δ x

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Lecture 5 Purdue University, Physics 220 4 Pulley Example Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. 1 2 1) T - m 1 g = m 1 a 1 2) T - m 2 g = -m 2 a 1 using a 1 = -a 2 2) T = m 2 g -m 2 a 1 1) m 2 g -m 2 a 1 - m 1 g = m 1 a 1 a 1 = (m 2 – m 1 )g / (m 1 +m 2 ) 1 T m 1 g 2 T m 2 g Compare the acceleration of boxes 1 and 2 A) |a 1 | > |a 2 | B) |a 1 | = |a 2 | C) |a 1 | < |a 2 | y x
Lecture 5 Purdue University, Physics 220 5 Pulley Example Two boxes are connected by a string over a frictionless pulley. Box 1 has mass 1.5 kg, box 2 has a mass of 2.5 kg. Box 2 starts from rest 0.8 meters above the table, how long does it take to hit the table. 1 2 1 T m 1 g 2 T m 2 g y x a 1 = (m 2 – m 1 )g / (m 1 +m 2 ) a = 2.45 m/s 2 Δ x = v 0 t + ½ a t 2 Δ x = ½ a t 2 t = sqrt(2 Δ x/a) t = 0.81 seconds

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Lecture 5 Purdue University, Physics 220 6 Kinematics Example • A leopard starts from rest at t=0 and runs in a straight line with a constant acceleration until t=3.0 s. Find the distance covered by the leopard between t=1.0 s and t=2.0 s as multiple of the distance covered between t=0 and t=1s
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Lecture05 - PHYSICS 220 Lecture 05 Free Fall and Apparent...

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