520 likes | 715 Views
Midterm 2 will be held on November 8. Covers units 4-9. Classical Mechanics Friction: Examples Work &Kinetic Energy. JFB Rotunda David Matthew Stephens Lei Shan (Mike). Schedule. Schedule. Schedule. Coupled Blocks with Friction.
E N D
Midterm 2 will be held on November 8. Covers units 4-9 Classical Mechanics Friction: ExamplesWork &Kinetic Energy JFB Rotunda David Matthew Stephens Lei Shan (Mike)
Coupled Blocks with Friction A block (m2)slides on a table pulled by a string attached to a mass (m1)hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string? m2 g m1
Tension in String • A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string? T1 T2 m2 g What is the relationship between the magnitude of the tension of the string at block 2 and the magnitude of the tension in the string at block 1? A) T1 > T2B)T1 = T2 C) T1 < T2 m1
Acceleration of coupled blocks A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string? m2 g m1 What is the relationship between the magnitudes of the acceleration of the two blocks? A) a1 = a2B) a1 < a2 C) a1 > a2
m2 1) FBD N T m2 g f T m1 m1 m2g m1g
m2 1) FBD 2) SF=ma N T m2 g f T m1 m1 m2g N = m2g m1g m1g – T = m1a T – mm2g = m2a add m1g – mm2g = m1a + m2a m1g – mm2g a = m1 + m2
m2 1) FBD 2) SF=ma N T m2 g f T m1 m1 m2g m1g m1g – T = m1a m1g – mm2g T = m1g – m1a a = m1 + m2 T is smaller when a is bigger
Midterm 2 will be held on November 8. Covers units 4-9 Classical Mechanics Lecture 7 Today’s Concepts: Work & Kinetic Energy JFB Rotunda David Matthew Stephens Lei Shan (Mike)
Work-Kinetic Energy Theorem The work done by force F as it acts on an object that moves between positions r1 andr2is equal to the change in the object’s kinetic energy:
Work-Kinetic Energy Theorem But again…!!! The work done by force F as it acts on an object that moves between positions r1 andr2is equal to the change in the object’s kinetic energy:
Work –Kinetic Energy Theorem: Work on sliding box Forces that act perpendicular to displacement perform no work!!!