Problem#1
At 11:00 A.M. on September 7, 2001, more than 1 million British school children jumped up and down for one minute. The curriculum focus of the “Giant Jump” was on earthquakes, but it was integrated with many other topics, such as exercise, geography, cooperation, testing hypotheses, and setting world records. Children built their own seismographs, which registered local effects. (a) Find the mechanical energy released in the experiment. Assume that 1 050 000 children of average mass 36.0 kg jump twelve times each, raising their centers of mass by 25.0 cm each time and briefly resting between one jump and the next. The free-fall acceleration in Britain is 9.81 m/s2. (b) Most of the energy is converted very rapidly into internal energy within the bodies of the children and the floors of the school buildings. Of the energy that propagates into the ground, most produces high-frequency “microtremor” vibrations that are rapidly damped and cannot travel far. Assume that 0.01% of the energy is carried away by a long-range seismic wave. The magnitude of an earthquake on the Richter scale is given byM = (log E – 4.8)/1.5
where E is the seismic wave energy in joules. According to this model, what is the magnitude of the demonstration quake? (It did not register above background noise overseas or on the seismograph of the Wolverton Seismic Vault, Hampshire.)
Jawab:
(a) One child in one jump converts chemical energy into mechanical energy in the amount that her body has as gravitational energy at the top of her jump:
U = mgy = 36.0 kg(9,81 m/s2)(0.250 m) = 88.3 J
For all of the jumps of the children the energy is
12(1.05 x 106)(88.3 J) = 1.11 x 109 J
(b) The seismic energy is modeled as
E = (0.01/100)(1.11 x 109 J) = 1.11 x 105 J
making the Richter magnitude
M = (log E – 4.8)/1.5 = [log(1.11 x 105 – 4.8)/1.5
M = 0.2
Problem#2
A bead slides without friction around a loop-the-loop (Fig. 1). The bead is released from a height h = 3.50R. (a) What is its speed at point A? (b) How large is the normal force on it if its mass is 5.00 g?
Answer:
(a) its speed at point A, We use
Ui + Ki = Uf + Kf
mgh + 0 = mg(2R) + ½ mv2
g(3.50R) = 2gR + ½ v2
v = [3.00gR]1/2
(b) we use
∑F = mv2/R
n + mg = mv2/R
n = m(v2/R – g) = m[3.00gR/R – g]
n = 2.00mg = 2.00(5.00 x 10-3 kg)(9.81 m/s2) = 0.0890 N downward
Problem#3
Dave Johnson, the bronze medalist at the 1992 Olympic decathlon in Barcelona, leaves the ground at the high jump with vertical velocity component 6.00 m/s. How far does his center of mass move up as he makes the jump?
Answer:
From leaving ground to the highest point,
Ui + Ki = Uf + Kf
0 + ½ mv2 = 0 + mgy
½ (6.00 m/s)2 = (9.81 m/s2)y
y = 1.84 m
Problem#4
A glider of mass 0.150 kg moves on a horizontal frictionless air track. It is permanently attached to one end of a massless horizontal spring, which has a force constant of 10.0 N/m both for extension and for compression. The other end of the spring is fixed. The glider is moved to compress the spring by 0.180 m and then released from rest. Calculate the speed of the glider (a) at the point where it has moved 0.180 m from its starting point, so that the spring is momentarily exerting no force and (b) at the point where it has moved 0.250 m from its starting point.
Answer:
(a) we use
Ui + Ki = Uf + Kf
½ kxi2 + ½ mvi2 = ½ kxf2 + ½ mvf2
½ (10.0 N/m)(–0.180 m)2 = 0 + ½ (0.150 kg)vf2
vf = 1.47 m/s
(a) we use
Ui + Ki = Uf + Kf
½ kxi2 + ½ mvi2 = ½ kxf2 + ½ mvf2
½ (10.0 N/m)(–0.180 m)2 = 0 + ½ (0.150 kg)vf2 + ½ (10.0 N/m)(0.25 m – 0.180 m)2
0.162 J = ½ (0.150 kg)vf2 + 0.0245 J
vf = 1.35 m/s
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