Problem#1
Air moving at 11.0 m/s in a steady wind encounters a windmill of diameter 2.30 m and having an efficiency of 27.5%. The energy generated by the windmill is used to pump water from a well 35.0 m deep into a tank 2.30 m above the ground. At what rate in liters per minute can water be pumped into the tank?
Answer:
Efficiency = useful output energy/total input energy = useful output power/total input power
With useful output power is
Pout = mwatergy/t = ρwaterVwatergy/t
total input power is
Pin = ½ mairvair2/t = ½ ρairVair(vair2/t)
the volume of water is Vair = πr2l, then
Pin = ½ ρair(πr2l)(vair2/t) = ½ ρair(πr2(vair2l/t)
Pin = ½πr2ρairvair3
where A is the length of a cylinder of air passing through the mill and vw is the volume of water pumped in time t. We need inject negligible kinetic energy into the water because it starts and ends
at rest.
e = [ρwaterVwatergy/t]/[ ½πr2ρairvair3]
Vwater/t = [eπr2ρairvair3]/[2ρwatergy]
Vwater/t = [(0.275)π(1.15 m)2(1.20 kg/m3)(11.0 m/s)3]/[2(1000 kg/m3)(9.80 m/s2)(35.0 m)]
Vwater/t = 2.66 x 10-3 m3/s = 160 L/min
Problem#2
A 20.0-kg cannon ball is fired from a cannon with muzzle speed of 1 000 m/s at an angle of 37.0° with the horizontal. A second ball is fired at an angle of 90.0°. Use the conservation of energy principle to find (a) the maximum height reached by each ball and (b) the total mechanical energy at the maximum height for each ball. Let y = 0 at the cannon.
Answer:
(a) the maximum height reached by each ball, we use
Ki + Ugi = Kf + Ugf
½ mvi2 + 0 = ½ mvf2 + mgyf
with vi2 = vxi2 + vyi2 and vf2 = vxf2 + vyf2.
½ m(vxi2 + vyi2) = ½ m(vxf2 + vyf2) + mgyf
But, vyf = 0 (maximum height) and vxi = vxf, then
yf = vyi2/2g = (1000 m/s sin37.00)2/[2(9.80 m/s2)]
yf = 1.85 x 104 m
and for the second
yf = vyi2/2g = (1000 m/s)2/[2(9.80 m/s2)] = 5.10 x 104 m
(b) The total energy of each is constant with value
E = ½mv2 = ½ (20.0 kg)(1000 m/s)2 = 1.00 x 107 J
Problem#3
A 2.00-kg ball is attached to the bottom end of a length of fishline with a breaking strength of 10 lb (44.5 N). The top end of the fishline is held stationary. The ball is released from rest with the line taut and horizontal (θ = 90.0°). At what angle θ (measured from the vertical) will the fishline break?
Answer:
In the swing down to the breaking point, energy is conserved:
Ki + Ugi = Kf + Ugf
0 + mgrcosθ = ½ mv2 + 0
v2/r = 2gcosθ
at the breaking point consider radial forces
∑F = mv2/r
Tmax – mgcosθ = mv2/r = 2mgcosθ
Tmx = 3mgcosθ
cosθ = Tmax/3mg = (44.5 N)/{2(2.00 kg)(9.80 m/s2)}
θ = 40.80
Problem#4
A daredevil plans to bungee-jump from a balloon 65.0 m above a carnival midway (Figure 1). He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10.0 m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke’s force law. In a preliminary test, hanging at rest from a 5.00-m length of the cord, he finds
that his body weight stretches it by 1.50 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon. (a) What length of cord should he use? (b) What maximum acceleration will he experience?
Answer:
For a 5-m cord the spring constant is described by
F = kx
mg = kx = k(1.5 m)
For a longer cord of length L the stretch distance is longer so the spring constant is smaller in inverse proportion:
k = (5/L)(mg/1.5) = 3.33mg/L
we use energy is conserved:
Ki + Ugi + Usi = Kf + Ugf + Usf
0 + mgyi + 0 = 0 + mgyf + ½kxf2
mg(yi – yf) = ½ kxf2 = ½ (3.33mg/L)xf2
(yi – yf) = ½ (3.33/L)xf2
here, yi – yf = L + xf = 55.0 m, then
55.0 m = ½ (3.33/L)(55.0 m – L)2
55.0 mL = ½ (3.33)(55.0 m – L)2 = 5.04 x 103 m2 – 183 mL + 1.67L2
1.67L2 – 238L + 5.04 x 103 = 0
L = [238 ± {(238)2 – 4(1.67)(5.04 x 103)}1/2]/[2(1.67)]
L = 25.8 m
only the value of L less than 55 m is physical.
(b) xmax = xf = 55.0 m – L = 55.0 m – 25.8 m = 29.2 m and k = 3.33mg/L, then from
∑F = ma
kxmax – mg = ma
(3.33mg/L) – mg = ma
a = 2.77g = 27.1 m/s2
Problem#5
Review problem. The system shown in Figure 2 consists of a light inextensible cord, light frictionless pulleys, and blocks of equal mass. It is initially held at rest so that the blocks are at the same height above the ground. The blocks are then released. Find the speed of block A at the moment when the vertical separation of the blocks is h.
 |
| Fig.2 |
Answer:
When block B moves up by 1 cm, block A moves down by 2 cm and the separation becomes 3 cm. We then choose the final point to be when B has moved up by h/3 and has speed vA/2. Then A has
moved down 2h/3 and has speed vA:
KAi + KBi + Ugi = KAf + KBf + Ugf
0 + 0 + 0 = ½ mvA2 + ½ m(vA/2)2 + mgh/3 – 2mgh/3
mgh/3 = 5mvA2/8
vA = [8gh/15]1/2
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