Last Time: Chapter 6 Today: Chapter 7 Last Time Work done by non- constant forces Work and springs Power Examples Today Poten&al Energy of gravity and springs Forces and poten&al energy func&ons Energy diagrams Examples with PE and conserva&on of energy
Poten/al Energy Remember Kine&c energy is the energy associated with the mo&on of an object The faster the object is moving the higher its kine&c energy. Energy can be transformed from one type to another; cannot be created nor destroyed. Poten/al Energy An energy associated with an object s posi&on. It is a stored energy that depends on an object s posi&on rela&ve to a reference point. Two main types of poten&al energy: Gravita&onal Elas&c It can be defined for any conserva&ve (path independent) force.
Gravita/onal poten/al energy Gravity exerts a downward force on an object Poten&al energy associated with a body s weight and height above the ground gravita&onal poten&al energy If an object moves from a height y i to a lower height y f, then the work done by gravity W grav =! F!! s = "mg(y f " y i ) = mg(y i " y f ) We define the gravita&onal poten&al energy as U grav! mgy The work done by gravity then is W grav =!"U grav If y f > y i (goes up), ΔU grav > 0 (poten&al increases), and W grav < 0 Then (takes work to push it up, gravity does nega&ve work on the object) If y f < y i (goes down), ΔU grav < 0 (poten&al decreases), and W grav < 0 Then (gravity does posi&ve work to bring object down)
Prelecture ques/on 1 Two balls, one having twice the mass of the other, are released from the same height. Just before hi[ng the ground, the kine&c energies of the balls are K1 and K2 respec&vely. How do K1 and K2 compare? a) K 2 = K 1 b) K 2 = 2K 1 c) K 2 = 4K 1 T. S&egler 10/08/2013 Texas A&M University
Gravita/onal work and poten/al For example, a ball drops a height h: The work done on the ball by gravity is posi&ve! F = mg(! ĵ) " $ d s! = (! ĵ)ds # %$ W grav = ' h 0! F & d! s = mgh The change in the poten&al energy is nega&ve because now there is less energy stored in the ball ΔU grav = - mgh In order to li` the ball back to its original height takes work: you have to supply some energy in order to replenish the poten&al Gravity does nega&ve work (takes energy away): W grav = mg(! Poten&al increases:!u grav = mg(y f " y i ) = mgh ĵ)" h( ĵ) =!mgh To go down and then up the total work done by gravity is mgh- mgh = 0, and the overall change in poten&al is - mgh + mgh = 0. These quan&&es only depend on the ini&al and final posi&ons; i.e. path independent
Clicker Ques/on A piece of fruit falls straight down. As it falls, A. the gravita&onal force does posi&ve work on it and the gravita&onal poten&al energy increases. B. the gravita&onal force does posi&ve work on it and the gravita&onal poten&al energy decreases. C. the gravita&onal force does nega&ve work on it and the gravita&onal poten&al energy increases. D. the gravita&onal force does nega&ve work on it and the gravita&onal poten&al energy decreases.
Gravity and the work energy theorem If gravity is the only force doing the work, then the work- energy theorem tells us that: W tot =!K W tot = "!U grav # $%!K = "!U Subs&tu&ng in for ΔK and ΔU grav 1 2 mv 2 f! 1 2 mv 2 i =!(mgy f! mgy i ) mgy i + 1 2 mv 2 i = 1 2 mv 2 f + mgy f or K i +U grav,i = K f +U grav, f = E tot (E tot is the total mechanical energy of the system) This equa&on tells us that the total mechanical energy of the system is conserved E tot = K + U grav Constant
Checkpoint ques/on 1 Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired straight up, ball 2 is fired straight down, and ball 3 is fired horizontally. Rank in order from largest to smallest the speeds of the balls, v1, v2, and v3, just before each ball hits the ground. a) v1 > v2 > v3 b) v3 > v2 > v1 c) v2 > v3 > v1 d) v1 = v2 = v3
Example Conserva&on of energy A 72 kg man jumps into a pool from a 3.5 m plakorm above the water. Find his speed as he hits the water if a) He just holds his nose and drops in b) He jumps up with ini&al speed of 2.5 m/s upward c) If he runs horizontally off at 2.5 m/s Remember: K i +U i +W other = K f +U f U grav = mgy K = 1 2 mv2
Example Conserva&on of Energy A pendulum of mass m and length L is released from rest from a horizontal posi&on as shown. a) What is it s speed at an angle θ from the horizontal? b) What is the tension in the string at θ=90? m θ L
Clicker Ques/on As a rock slides from A to B along the inside of this fric&onless hemispherical bowl, mechanical energy is conserved. Why? (Ignore air resistance.) A. The bowl is hemispherical. B. The normal force is balanced by centrifugal force. C. The normal force is balanced by centripetal force. D. The normal force acts perpendicular to the bowl s surface. E. The rock s accelera&on is perpendicular to the bowl s surface.
Other forces that do work (like fric&on) Energy conserva&on so far has only considered the work done by gravity and its effect on the change in kine&c energy of an object. If other forces do work, we need to take those into account. If they are not posi&on dependent (i.e. non- conserva&ve) then we lump them together into W other Recall: W grav =!"U grav #% $ W tot = W other +W grav = "K&% W = W! "U = "K tot other grav K i +U grav,i +W other = K f +U grav, f If the other work is nega&ve (like fric&on), then it takes mechanical energy away from the object f i E tot < E tot If the other work is posi&ve (like an engine), it adds mechanical energy to the object E f tot > E i tot
Example Conserva&on of energy and work other We want to slide a 12 kg crate up a 2.5 m long ramp inclined at 30. Ignoring fric&on you calculate you can do this by giving it an ini&al speed of 5.0 m/s. But fric&on is not negligible, it only slides 1.6m up the ramp, stops, and slides back down. a) Find the magnitude of the fric&on force ac&ng on the crate, assuming its constant. b) How fast is the crate moving when it reaches the boqom? Remember: K i +U i +W other = K f +U f U grav = mgy U el = 1 2 kx2 K = 1 2 mv2 r! W! F " d r! 2 # r 1
Example Conserva&on of energy and work other cont. a) Find the magnitude of the fric&on force ac&ng on the crate, assuming its constant. b) How fast is the crate moving when it reaches the boqom?
Elas/c poten/al energy (springs) A body is elas<c if it returns to its original shape a`er being deformed. Elas<c poten<al energy is the energy stored in an elas&c body, such as a spring. If we stretch a spring, the spring does nega&ve work on the block When we go back to equilibrium the spring does posi&ve work on the block If we let W el =! 1 2 kx 2 f U el = 1 2 k(x! x 0 )2 x 0 x f Then, W el = 1 2 kx 2 f W el =!"U el x 0 x f
Prelecture ques/on 2 In Case 1, a mass M hangs from a ver&cal spring having spring constant k and is at rest in its equilibrium posi&on. In Case 2 the mass has been li`ed a distance D ver&cally upward. If we define the poten&al energy in Case 1 to be zero, what is the poten&al energy of Case 2? a) b) c) 1 2 kd2 MgD 1 2 kd2 + MgD
Checkpoint ques/on 2 A box sliding on a horizontal fric&onless surface runs into a fixed spring, compressing it a distance x 1 from its relaxed posi&on while momentarily coming to rest. If the ini&al speed of the box were doubled, how far x 2 would the spring compress? a) x 2 = (2) x 1 b) x 2 = 2 x 1 c) x 2 = 4 x 1
Clicker Ques/on A block aqached to a spring is oscilla&ng between point x (fully compressed) and point y (fully stretched). The spring is un- stretched at point o. At point o, which of the following quan&&es is at its maximum value? x o y a) The block s kine&c energy b) The spring s poten&al energy c) Both a) and b) d) None of the above
Example Springs and conserva&on of energy A 6.0kg box moving at 3.0m/s on a smooth horizontal surface runs into a light spring with a force constant of 75 N/cm. Use the work- energy theorem to find the maximum compression of the spring. v i Δx Remember: K i +U i +W other = K f +U f U el = 1 2 kx2 K = 1 2 mv2
Clicker Ques/on You toss a 0.150- kg baseball straight upward so that it leaves your hand moving at 20.0 m/s. The ball reaches a maximum height y 2. What is the speed of the ball when it is at a height of y 2 /2? Ignore air resistance. v 2 = 0 y 2 A. 10.0 m/s B. less than 10.0 m/s but greater than zero C. greater than 10.0 m/s D. not enough informa&on given to decide m = 0.150 kg v 1 = 20.0 m/s y 1 = 0