C6.87
No friction
C
A
B
Initial mechanical energy the same for the two blocks.
Block at left reaches highest point at A, at y = H.
Block at right leaves shorter plane at, B, at y = H1
goes on to highest point at C, at y = H1 + H2.
Is A higher than C?
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v=0
C
A
B
At A: mechanical energy entirely potential, as block has come to rest.
At C: mechanical energy is a mixture of potential and kinetic, as the
block is still moving to the right.
So E = mgH = mg(H1 + H2) + KEC
Therefore, H > H1 + H2
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6.42
Skier starts from rest
A
B
v
r = 36 m
Negligible friction and air resistance
What must be the height h if the skier just loses contact with the
snow at the crest of the second hill at B?
Mechanical energy is conserved, so EA = EB.
That is, 0 + mgh = mv2/2 + 0
So, v2 = 2gh
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v2 = 2gh
FN
B
mg
r = 36 m
So, v2 = 2gh
If skier loses contact with snow at B, then FN = 0
So, centripetal acceleration toward centre of curved slope of radius r
is provided only by the skier’s weight, mg.
That is, mg = mv2/r and v2 = rg
But v2 = 2gh, so 2gh = rg and h = r/2 = 18 m
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6.44
v B
v
mg + FN
A
Initial speed of model car = 4 m/s. What is maximum radius, r, if
the car is to remain in contact with the circular track? No friction.
Conservation of mechanical energy:
EA = EB
Or, KEA + PEA = KEB + PEB
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v B
mg + FN
Car loses contact if
FN = 0
v0 = 4 m/s
A
KEA + PEA = KEB + PEB
That is, mv20/2 + 0 = mv2/2 + mgh
= mv2/2 + 2mgr
→ v2 = v20 − 4rg
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v B
→ v2 = v20 − 4rg
mg + FN
Car loses contact if
FN = 0
A
Centripetal force at B: mv2/r = mg + FN
If FN = 0, v2 = rg
So v2 = rg = v20 − 4rg
v20 = 5rg → r = v20/5g = 42/5g = 0.32 m
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Non-conservative Forces
The work done by a non-conservative force depends on the path.
Friction – the longer the path taken, the more the (negative)
work done by the friction force.
Defining a potential energy relies on a conservative force so that
the work done in moving an object from A to B depends only on
the positions of A and B.
Examples of non-conservative forces
• Static and kinetic friction forces
• Air resistance
• Tension, or any applied force
• Normal force
• Propulsion force in a rocket
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Work-energy theorem revisited
y=h
A few pages back –
Work done by applied force
F
a=0
y=0
mg
= (change in KE) + (change in PE)
So W = Fh = ΔKE + ΔPE
Identify W with the work done by the external
force, F, which is non-conservative:
So, Wnc = ΔKE + ΔPE
Work-energy theorem extended to non-conservative forces
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6.46
A 0.6 kg ball is pitched from a height of 2 m above the
ground at 7.2 m/s. The ball travels at 4.2 m/s when it is 3.1
m above the ground.
How much work is done by air resistance, a non-conservative force?
Wnc = !KE + !PE (work-energy theorem)
= m(v2f − v20)/2 + mg(y f − y0)
= 0.6(4.22 − 7.22)/2 + 0.6g(3.1 − 2) = −14.1 J
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6.48
A projectile of mass 0.75 kg is shot straight up with an
initial speed of 18 m/s.
a) How high will it go if there is no air resistance?
B
v = 0, y = h
A
vo = 18 m/s, yo = 0
No air resistance, so energy is conserved and EA = EB
So, mv20/2 + 0 = 0 + mgh
v20 182
and, h =
=
= 16.5 m
2g
2g
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b) If the projectile rises to only 11.8 m, find the average force due
to air resistance.
Work-energy theorem: Wnc = !KE + !PE
!KE = 0 − mv20/2 = −(0.75 kg) × (18 m/s)2/2 = −121.5 J
(final - initial)
!PE = mgh − mgy0 = (0.75 kg) × 9.8 × (11.8 − 0 m) = 86.7 J
(final - initial)
So that Wnc = −121.5 + 86.7 = −34.8 J
Wnc = F × h, where F = force of air resistance
Therefore F = Wnc/h = (−34.8 J)/(11.8 m) = −2.95 N
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Power
Power is the rate of doing work, or the rate of change of
mechanical energy due to an applied force.
Unit: 1 watt (W) = 1 J/s
W Fs
s
Power, P = =
= F × = Fv
t
t
t
P = Fv
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Power
A 1,100 kg car is accelerated from rest at 4.6 m/s2 for 5 s.
Find the average power generated by the force accelerating it.
Applied force, F = ma = (1,100 kg)×(4.6 m/s2) = 5060 N
Average power, P̄ = F v̄
Average speed, v̄ = (0 + at)/2 = 4.6 × 5/2 = 11.5 m/s
So P̄ = (5060 N) × (11.5 m/s) = 58, 190 W = 58.2 kW
42