November 28, 2007
PHY2053
Quiz 9 (Chapter 9.1-9.8)
Name:
UFID:
1. (5 pts) A steel cable 1.50 m long and 3×10 m² in cross sectional area hangs from the
ceiling of an elevator and 800-kg mass is attached the end of the cable. The cable is
stretched by 2.50 mm when the elevator is accelerating upward. Find the acceleration of
the elevator. (Young’s modulus of steel is 20×10¹º Pa.)
The tension in the cable is given by
F = AY(ΔL/L) = 3×10 ×20×10¹º×(2.5×10ˉ³/1.5) = 10000 N
Applying the Newton’s second law to the mass, we get,
ma = F-mg ⇒ a = (F-mg)/m = (10000-800×9.8)/800 = 2.70 m/s
2. (5 pts) A light spring with a spring constant of 150 N/m rests vertically on the bottom
of a beaker of water (density = 1000 kg/m³). A 3.00 kg aluminum block (density = 2700
kg/m³) is attached to the end of the spring. What is the compression ΔL of the spring?
Applying the equilibrium condition, we get
B+kΔL-mg = 0 ⇒ ΔL = (mg-B)/k
The buoyant force is given by
B = ρVg = ρ(m/ρ’)g, where ρ = 1000 kg/m³ and ρ’ =2700 kg/m³.
Plugging this into the first equation, we get
ΔL = mg(1-ρ/ρ’)/k = 3×9.8(1-1000/2700)/150 = 0.123 m.
3. (5 pts) A water tank open to the atmosphere at the top has two small holes in its side,
one above the other. The holes are 4.00 cm and 12.00 cm above the floor. How high does
water stand in the tank if the two streams of water hit the floor at the same place?
Let the height of the water, the top hole and the bottom hole be Y, y and y’. The
velocities at the holes are given by Bernoulli’s law:
(1/2)mv²+mgy = mgY ⇒ v = √(2g(Y-y)) and (1/2)mv²+mgy’ = mgY ⇒ v’ = √(2g(Y-y’))
We calculate the time when the ejected water streams reach the floor and then the
position where the streams hit the floor:
y = (1/2)gt² ⇒ t = √(2y/g), d = vt = √(2g(Y-y)) √(2y/g)
y’ = (1/2)gt’² ⇒ t’ = √(2y’/g), d’ = v’t’ = √(2g(Y-y’)) √(2y’/g)
Since d = d’, we get
√(2g(Y-y)) √(2y/g) = √(2g(Y-y’)) √(2y’/g) ⇒ (Y-y)y = (Y-y’)y’
⇒ Y(y-y’) = y²-y² ⇒ Y = y+y’ = 4+12 = 16.0 cm.
4. (5 pts) Water flowing in a horizontal pipe is at a pressure of 2.0×10 Pa at a point
where its cross-sectional area is 1.00 m². When the pipe narrows to 0.500 m², the
pressure drops to 1.5×10 Pa. Find the speed of water in the wider pipe.
Applying the equation of continuity we get
vA = v’A’ ⇒ v’ = (A/A’)v = (1/0.5)v = 2v
Plugging this into the Bernoulli’s equation, we get
P+(1/2)ρv² = P’+(1/2)ρv’² ⇒P+(1/2)ρv² = P’+(1/2)ρ(2v)² ⇒ (3/2)ρv² = P-P’
⇒v = √(2(P-P’)/3ρ) = √(2(2-1.5)×10 /(3×1000)) = 5.77 m/s