July 5, 2008
PHY2053 Discussion
Quiz 5 (Chapter 5)
Name:
UFID:
**1. (5pts) Three blocks with masses of 4.00 kg, 6.00 kg and 12.0 kg are attached by
strings over frictionless pulleys: The 6-kg block is on a horizontal table with a coefficient
of kinetic friction of 0.250. Two strings are attached to the block, one of which passes
over a pulley at the left edge of the table and connected to the 4-kg hanging block, and
the other string passes over a pulley at the right edge of the table and connected to
12-kg hanging block. Assuming the system starts from rest, find the speed of the 12-kg
block when it falls down a distance of 3.00 m
Since the three blocks are connected together, when the 12-kg block falls down a
distance Δx = 3 m, the 4-kg rises 3 m up and 6-kg block slides 3 m on the table. They
also have a common velocity v. Applying the work-energy theorem to the system, we
have
Wnc = ΔE ⇒ -μm2gΔx = (1/2)(m1+m2+m3)v²+m1gΔx-m3gΔx
⇒ (1/2)(m1+m2+m3)v² = (m3-m1-μm2)gΔx
⇒ v = √[{2(m3-m1-μm2)gΔx}/( m1+m2+m3)] = √{2×(12-4-0.25×6)×9.8×3/22} = 4.17 m/s
*2. (5pts) A launching mechanism of a toy gun consists of a spring of unknown spring
constant. If its spring is compressed a distance of 0.160 m and the gun fired vertically,
the gun can launch a 30.0 g projectile from rest to a maximum height of 12.0 m above
the starting point of the projectile motion. What is the spring constant?
A projectile rises a distance Δy1 = 0.16 m before it leaves the spring and then rises Δy2 =
12 m to reach the maximum height. Applying the energy conservation law to the initial
and final positions, we get
(1/2)kΔy1² = mg(Δy1+Δy2)
⇒ k = 2 mg(Δy1+Δy2)/ Δy1² = 2×0.03×9.8×12.16/0.16² = 279 N/m
***3. (5pts) A 500-g block is shot from an 800 N/m horizontal spring. The spring is
compressed a distance of 0.400 m and released. After leaving the spring, the block slides
0.600 m on the horizontal ground, slides 1.00 m up a 30.0˚ ramp, and then leaves the
ramp and undergoes a projectile motion. The horizontal ground and the ramp have a
coefficient of kinetic friction of 0.200. What is the maximum height of the projectile
motion, measured from the ground level?
To calculate the maximum height, we need the initial velocity of the projectile motion.
Before the block leaves the lamp, it slides Δx1 =1 m (= 0.4+0.6 ) on the horizontal ground
and slides Δx2 = 1m up the incline. Applying the work-energy theorem, we have
-μmgΔx1-μmgcosθΔx2 = (1/2)mv²+mgΔx2sinθ-(1/2)kd²
⇒ (1/2)mv² = (1/2)kd²-μmg(Δx1+Δx2cosθ)
v = √{kd²/m-2μmg(Δx1+Δx2cosθ)}
= √800×0.4²/0.5-2×0.2×0.5×9.8×(1+1×cos30˚)} = 15.9 m/s
At the peak of the projectile motion, the vertical component of the velocity is zero. Thus,
the velocity at the peak is vx = vcosθ. Applying the energy conservation, we have
(1/2)mv² = (1/2)m(vcosθ)² + mgΔy ⇒ mgΔy = (1/2)m(vsinθ)²
⇒ Δy = (1/2g)(vsinθ)² = 3.19 m
Adding the height of the ramp h = Δx2sinθ = 1×sin30˚ = 0.5 m, the maximum height is
ymax = Δy+h = 3.19+0.5 = 3.69 m
**4. (5pts) 1.20×103-kg car starts from rest, accelerates uniformly to a speed of 16.0 m/s
in 8.00 s and then continues at constant speed. Assuming that air resistance is constant
at 500 N during the whole motion, find the ratio of the power just before the car stops
accelerating to the power just after stopping the acceleration.
The acceleration of the car is
a = Δv/Δt = 16/8 = 2 m/s².
Before the car stops the acceleration, the force exerted by the engine is
ma = Fb-R ⇒ Fb = R+ma = 500+1200×2 = 2900 N
After the car stops the acceleration, the force exerted by the engine is
0 = Fa-R ⇒ Fa = R = 500 N
The speed of the car just before and after stopping acceleration is 16 m/s. The ratio of
the power is given by
Pb/Pa = Fbv/Fav = Fb/Fa = 2900/500 = 5.80