Download Chapter 7 Conserva)on of Energy (cont`d)

Chapter 7 Conserva.on of Energy (cont’d) Mechanical energy conserva.on Examples Work by non‐conserva.ve forces March 4, 2010 Recollect: Conserva.on of Energy •  If only conserva.ve forces are present, the total kine.c plus poten.al energy of a system is conserved.  Mechanical Energy = Poten.al Energy (U) + Kine.c Energy (K) Conserva.ve forces interchange U  K (work done), but E = K + U is a constant. ΔE = ΔK + ΔU = 0 Work‐kine.c energy theorem: ΔK = W thus ΔU = ‐W, for conserva.ve forces only. •  Both K and U can change, but E = K + U remains constant. •  But, if non‐conserva.ve forces act, then energy can be dissipated in other forms (heat, for example) Example: The simple pendulum •  Suppose we release a mass m from rest a height h1 above its lowest possible eleva.on. Assuming no fric.on (air drag): –  What is the maximum speed of the mass and where does this happen? –  To what eleva.on h2 does it rise on the other side? m
h1
h2
v
Example: The simple pendulum •  Kine.c + poten.al energy is conserved since gravity is a conserva.ve force (E = K + U is a constant) •  Choose y=0 at the boWom of the swing, and set the constant U=0 at y=0 Thus the general energy func.on of v, y: E = ½mv2 + mgy m
h1
h2
v
Example: The simple pendulum •  E = ½mv2 + mgy –  Ini.ally, y = h1 and v = 0, so E = mgh1. –  Since E = mgh1 ini.ally, and energy is conserved, E = mgh1 at all .mes. y
m
h1
y=0
Example: The simple pendulum E = ½mv2 + mgy Velocity is maximum where poten.al energy is lowest, at boWom of the swing. So, at y = 0, ½mv2 = mgh1  v2 = 2gh1  v = (2gh1)1/2 y
m
h1
y=0
v
Example: The simple pendulum •  To find maximum eleva.on on other side, note that maximum is reached when v=0. Since E = mgh1, and maximum poten.al energy on right is mgh2, h2=h1. The ball returns to its original height. y
m
h1
y=0
h2
Example: Airtrack and Glider •  A glider of mass M is ini.ally at rest on a horizontal fric.onless track. A mass m is aWached to it with a massless string hung over a massless pulley as shown. What is the speed v of M a^er m has fallen a distance d? v
M
m
d
v
Example: Airtrack and Glider •  Kine.c + poten.al energy is conserved since all forces are conserva.ve. •  Choose ini.al configura.on to have U=0. ΔK = ‐ΔU 2mgd
1
2
( m + M )v = mgd ⇒ v =
2
m+M
v
M
m
d
v
Problem: Toy car •  A toy car slides on a fric.onless track shown below. It starts at rest, drops a height d, moves horizontally at speed v1, rises a height h, and ends up moving horizontally with speed v2. –  Find v1 and v2. v2
d
v1
h
Problem: Toy car •  K+U is conserved, so ΔK = ‐ΔU •  When the eleva.on decreases a distance D, ΔU = ‐mgd, ΔK = ½mv12. •  Solving for the speed: v1 = 2gd
€
d
v2 = 2g(d − h)
v2
€
v1
h
A projec.le of mass m is propelled from ground level with an ini.al kine.c energy of 450 J. At the exact top of its trajectory, its kine.c energy is 250 J. To what height, in meters, above the star.ng point does the projec.le rise? Assume air resistance is negligible. A)  50/(mg)
B)  250/(mg)
C)  700/(mg)
E)  350/(mg)
A projec.le of mass m is propelled from ground level with an ini.al kine.c energy of 450 J. At the exact top of its trajectory, its kine.c energy is 250 J. To what height, in meters, above the star.ng point does the projec.le rise? Assume air resistance is negligible. A)  50/(mg)
B)  250/(mg)
C)  700/(mg)
E)  350/(mg)
K1 + U1 = K 2 + U 2 , with U i =mgh i
So mg(h 2 -h1 )=K1 -K 2
h 2 -h1 =(K1 -K 2 )/(mg)
Ques.on •  A box sliding on a horizontal fric.onless surface runs into a fixed spring, compressing it to a distance x1 from its relaxed posi.on while momentarily coming to rest. –  If the ini.al speed of the box were doubled and its mass were halved, what would be the distance x2 that the spring would compress? A) x2=x1 B) x2=x1√2 C) x2=2x1 x
Ques.on •  A box sliding on a horizontal fric.onless surface runs into a fixed spring, compressing it to a distance x1 from its relaxed posi.on while momentarily coming to rest. –  If the ini.al speed of the box were doubled and its mass were halved, what would be the distance x2 that the spring would compress? A) x2=x1 B) x2=x1√2 C) x2=2x1 (solution on next slide)
x
Solu.on Again, use the fact that WNET = ΔK. Here, WNET = WSPRING = ‐½kx2 and ΔK = ‐½mv2  kx2 = mv2, and thus v = x(k/m)1/2, x = v(m/k)1/2 So doubling v and halving m increases x by a factor of √2. x
Non‐conserva.ve forces: •  If the work done does not depend on the path taken, the force is said to be conserva.ve. •  If the work done does depend on the path taken, the force is said to be non‐conserva.ve. •  An example of a non‐conserva.ve force is fric.on. –  When pushing a box across the floor, the amount of work that is done by fric.on depends on the path taken. •  Work done is propor.onal to the length of the path! –  The mechanical energy is converted to heat. Spring pulls on mass: with fric.on •  Suppose spring pulls on block, but now there is a nonzero coefficient of fric.on μ between the block and the floor. •  The total work done on the block is now the sum of the work done by the spring, Ws (same as before), and the work done by the fric.on Wf, W = f ⋅ Δr = −µmgd
f
(not related to either kine.c energy or poten.al energy) Work‐energy theorem now reads: Wnet = Wcons + Wf = ΔK €
m
stretched position (at rest)
d
m
vr
f = µmg
relaxed position
€
Spring pulls on mass: with fric.on •  Again use Wnet = Ws + Wf = ΔK 1 2
1 2
Wf = µmgd
Ws = kd
ΔK = mvr
2
2
1 2
1 2
kd − µmgd = mvr
2
2
k 2
vr =
d − 2µgd
m
€
€
€
m
stretched position (at rest)
d
m
vr
f = µmg
relaxed position
Many forms of energy: Ques.on Which statement is true? A. Mechanical energy is always conserved B. Total energy is always conserved C. Poten.al energy is always conserved Ques.on Which statement is true? A. Mechanical energy is always conserved B. Total energy is always conserved C. Poten.al energy is always conserved Energy of one form can be converted to
another form, but the total energy remains
the same.
Conserva.on of energy is more broadly applicable than Newton’s laws •  Newton’s laws do not apply to systems that are fast‐
moving close to the speed of light (where Einstein’s theory of rela.vity applies) and to very small systems (where quantum mechanics applies), but conserva.on of energy is always valid. Conserva.on laws are the consequence of symmetries: Energy conserva.on  Time transla.on invariance.