PHY183-Lecture16

# PHY183-Lecture16 - Clicker Quiz A ball of mass m is thrown...

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1 February 9, 2012 Physics for Scientists&Engineers 1 1 Physics for Scientists & Engineers 1 Spring Semester 2012 Lecture 16 Conservation of Mechanical Energy and Non-conservative Forces Clicker Quiz A ball of mass m is thrown vertically into the air with an initial speed v . Which of the following equations correctly describes the maximum height, h , of the ball? February 9, 2012 Physics for Scientists&Engineers 1 2 A) h = v 2 g B) h = g 1 2 v 2 C) h = 2 mv g D) h = mv 2 g E) h = v 2 2 g 0 0 2 0 2 0 1 0 0 2 2 K U K U mgh mv v h g + = + + = + = February 9, 2012 Physics for Scientists&Engineers 1 3 Defending the Castle Defend the castle from attackers! Shoot rocks from catapult with launch speed of 14.2 m/s, from the courtyard over the castle walls onto the attackers in front of the castle, at elevation 7.2 m below courtyard. Question: What is the speed with which your rocks hit the attackers? Answer: Hard way: find appropriate launch angle to clear wall; decompose initial velocity vector into components; solve for y -component of velocity as a function of time or altitude, following trajectory to impact; take square root of sum of the squares of the velocity components Defending the Castle THINK Much easier way is to use energy concepts Once the rock is launched, only the conservative force of gravity is acting on the rock Total mechanical energy is conserved SKETCH Define initial and final conditions RESEARCH Conservation of mechanical energy gives us February 9, 2012 Physics for Scientists&Engineers 1 4 E = K + U = K 0 + U 0 Defending the Castle The kinetic energy of the projectile is The potential energy of the projectile is SIMPLIFY Substitute K and U in E The mass of rock cancels out February 9, 2012 Physics for Scientists&Engineers 1 5 K = 1 2 mv 2 U = mgy E = 1 2 mv 2 + mgy = 1 2 mv 0 2 + mgy 0 1 2 v 2 + gy = 1 2 v 0 2 + gy 0 Defending the Castle Solve for the speed when the rock lands in the courtyard CALCULATE y y 0 = 7.20 m and v 0 = 14.2 m/s, so ROUND DOUBLE-CHECK The speed when the rock hits the ground is 18.5 m/s, which is faster than the initial speed of 14.2 m/s Note that our answer is independent of θ 0 February 9, 2012 Physics for Scientists&Engineers 1 6 v = v 0 2 + 2 g y y 0 ( ) v = 14.2 m/s ( ) 2 + 2 9.81 m/s 2 ( ) 7.20 m ( ) = 18.517667 m/s v = 18.5 m/s

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2 February 9, 2012 Physics for Scientists&Engineers 1 7 Race of Two Balls Revisited Two balls on different tracks are released at the same time Ball 1 rolls on a horizontal track Ball 2 rolls on a “roller coaster” track We saw the ball on the “roller coaster” track finished first Which ball will have the higher speed when it reaches the end of the track? February 9, 2012 Physics for Scientists&Engineers 1 8 Clicker Question Choose the correct answer: A) Ball 1(horizontal track) will have a higher speed when it reaches the end of the track B) Ball 2 will have a higher speed when it reaches the end of the track C) Both balls will have the same speed when they reach the end of the track Sliding Chain PROBLEM A uniform chain of total mass m
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