Sect. 8-3: Mechanical Energy & It’s Conservation

GENERALLY In any process, total energy is neither created nor destroyed. The total energy of a system can be transformed from one form to another & from one object to another, but the total amount remains constant  The Law of Conservation of Total Energy

In Ch. 7, we learned the Work-Energy Principle: The net work W done on an object = the change in the object’s kinetic energy K W net =  K  The Work-Energy Principle (As we’ve said, this is Newton’s 2 nd Law in Work-Energy language!) Note: W net = The work done by the net (total) force.

If several forces act (both conservative & non-conservative), The total work done is: W net = W C + W NC W C = the work done by all conservative forces W NC = the work done by all non-conservative forces The Work-Energy Principle still holds: W net =  K For conservative forces (by the definition of potential energy U!): W C = -  U   K = -  U + W NC or: W NC =  K +  U

 So, in general, W NC =  K +  U The Total Work done by all Non-conservative forces = The total change in K + the total change in U

In general, we just found that: W NC =  K +  U In the (Very!) SPECIAL CASE where there are CONSERVATIVE FORCES ONLY!  W NC = 0 or  K +  U = 0  the Principle of Conservation of Mechanical Energy Note: This ISN’T (quite) the same as the Law of Conservation of (total) Energy! It’s a special case of this law (where all forces are conservative) Sect. 8-3: Mechanical Energy & Its Conservation

For the SPECIAL CASE of CONSERVATIVE FORCES ONLY!  K +  U = 0  the Principle of Conservation of Mechanical Energy Note Again! This is NOT (quite) the same as the Law of Conservation of (total) Energy! The Principle of Conservation of Mechanical Energy is a special case of this law (where all forces are conservative)

For CONSERVATIVE FORCES ONLY!  K +  U = 0  Principle of Conservation of Mechanical Energy Note! This came from the Work-Energy Principle & the definition of Potential Energy for conservative forces! Since the Work-Energy Principle is N’s 2 nd Law in Work-Energy language, & since Conservation of Mechanical Energy came from this principle, Conservation of Mechanical Energy is Newton’s 2 nd Law in Work-Energy language, in the special case of Conservative Forces only!

Conservation of Mechanical Energy So, for conservative forces ONLY! In any process  K +  U = 0 Conservation of Mechanical Energy Definition of Mechanical Energy: E  K + U Conservation of Mechanical Energy  In any process with only conservative forces acting,  E = 0 =  K +  U or E = K + U = Constant So, in any process with only conservative forces acting, the sum K + U is unchanged The mechanical energy changes from U to K or K to U, but the sum remains constant.

If & ONLY if there are no nonconservative forces, the sum of the kinetic energy change  K & the potential energy change  U in any process is zero. The kinetic energy change  K & the potential energy change  U are equal in size but opposite in sign. The Total Mechanical Energy is defined as: Conservation of Mechanical Energy has the form:

Summary: The Principle of Conservation of Mechanical Energy If only conservative forces are acting, the total mechanical energy of a system neither increases nor decreases in any process. It stays constant; it is conserved.

Conservation of Mechanical Energy   K +  U = 0 or E = K + U = Constant For conservative forces ONLY (gravity, spring, etc.) Suppose, initially: E = K 1 + U 1 & finally: E = K 2 + U 2 E = Constant  K 1 + U 1 = K 2 + U 2 A very powerful method of calculation!!

Conservation of Mechanical Energy   K +  U = 0 or E = K + U = Constant For gravitational U: U grav = mgy E = K 1 + U 1 = K 2 + U 2  (½)m(v 1 ) 2 + mgy 1 = (½)m(v 2 ) 2 + mgy 2 y 1 = Initial height, v 1 = Initial velocity y 2 = Final height, v 2 = Final velocity

 U 1 = mgh, K 1 = 0  U 2 = 0, K 2 = (½)mv 2 K + U = same as at points 1 & 2 the sum K + U remains constant K 1 + U 1 = K 2 + U mgh = (½)mv v 2 = 2gh all U half K half U all K

Sect. 8-4: Using Conservation of Mechanical Energy Original height of the rock: y 1 = h = 3.0 m S Speed when it falls to l y 3 = 1.0 m a above the ground? K 1 + U 1 = K 2 + U 2 = K 3 + U mgh = (½)mv = (½)m(v 3 ) 2 + mgy 3 Example 8-3: Falling Rock U 3 = mgy 3 K 3 = (½)m(v 3 ) 2 U 1 = mgy 1, K 1 = 0 U 2 = 0, K 2 = (½)m(v 2 ) 2

Example 8-3 What is it’s speed at y = 1.0 m? Conservation of Mechanical Energy!  (½)m(v 1 ) 2 + mgy 1 = (½)m(v 2 ) 2 + mgy 2 (The mass cancels in the equation!) y 1 = 3.0 m, v 1 = 0 y 2 = 1.0 m, v 2 = ? Gives: v 2 = 6.3 m/s  U only  Part U, part K  K only

Example – Free Fall A ball is dropped from height h above the ground. Calculate the ball’s speed at distance y above the ground. Neglecting air resistance, the only force is gravity, which is conservative. So, we can use Conservation of Mechanical Energy y 1 = h U 1 = mgh K 1 = 0 y 2 = y U 2 = mgy K 2 = (½)mv 2 y v v y 3 = 0, U 3 = 0 K 3 = (½)m(v 3 ) 2 v

Conservation of Mechanical Energy K 2 + U 2 = K 1 + U 1 In general, for an initial velocity v 1, solve for v: The result for v is, of course, consistent with the results from the kinematics of Ch. 2. For this specific problem, v 1 = 0 & K 1 = 0, since the ball is dropped. v = (v 1 ) 2 + 2g(h – y)

Example 8-4: Roller Coaster Mechanical energy conservation! (Frictionless!)  (½)m(v 1 ) 2 + mgy 1 = (½)m(v 2 ) 2 + mgy 2 (mass cancels!) Only height differences matter! Horizontal distance doesn’t matter! Speed at the bottom? y 1 = 40 m, v 1 = 0 y 2 = 0 m, v 2 = ? Gives: v 2 = 28 m/s For what y is v 3 = 14 m/s? Use: (½)m(v 2 ) = (½)m(v 3 ) 2 + mgy 3 Gives: y 3 = 30 m Height of hill = 40 m. Starts from rest at the top. Calculate: a. The speed of the car at the hill bottom. b. The height at which it has half this speed. Ii Take y = 0 at the bottom of the hill.

Conceptual Example 8-5: Speeds on 2 Water Slides Two water slides at a pool are shaped differently, but start lllll at the same height h. Two riders, Paul & Kathleen, start from rest at the same time on different slides. Ignore friction & assume that both slides have the same path length. a. Which rider is traveling FF faster at the bottom? b. Which rider makes it to the FF bottom first? WHY?? Demonstration! frictionless water slides! both start here! 

An actor, mass m actor = 65 kg, in a play is to “fly” down to stage during performance. Harness attached by steel cable, over 2 frictionless pulleys, to sandbag, mass m bag = 130 kg, as in figure. Need length R = 3 m of cable between nearest pulley & actor so pulley can be hidden behind stage. For this to work, sandbag can never lift above floor as actor swings to floor. Let initial angle cable makes with vertical be θ. Calculate the maximum value θ can have such that sandbag doesn’t lift off floor. Example – Grand Entrance

Free Body Diagrams  Actor Sandbag at bottom Step 1: To find actor’s speed at bottom, let y i = initial height above floor & use Conservationof Mechanical Energy K i + U i = K f + U f or 0 + m actor gy i = (½)m actor (v f ) (1) mass cancels. From diagram, y i = R(1 – cos θ) So, (1) becomes: (v f ) 2 = 2gR(1 – cosθ) (2) Step 2: Use N’s 2 nd Law for actor at bottom of path (T = cable tension). Actor: ∑F y = T – m actor g = m actor [(v f ) 2 /R] or T = m actor g + m actor [(v f ) 2 /R] (3) Step 3: Want sandbag to not move.  N’s 2 nd Law for sandbag: ∑F y = T – m bag g = 0 or T = m bag g (4) Combine (2), (3), (4) : m bag g = m actor g + m actor [2g(1 – cosθ)]. Solve for θ: cosθ = [(3m actor - m bag )/(2 m actor )] = 0.5 or, θ = 60 °