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Unformatted text preview: Physics 171.101 Exam 2 Answers
Prof. Barnett March 16, 2006 You may use anything written on the front and back of ONE 3”x5” index card.
Everything should be written in ink and no “snow paint” should be used.
Show your work. Don’t just write down an answer. Use 1. Bush, MB 2 75 kg, and Cheney, MC 2 100 kg, g = 10 m/sec2, G : 6.7 X 10‘11N-m2/kg2 100 kg are on a dude ranch in Crawford, Texas. They
have roped a calf, Torino, and are trying to drag
it to a ﬁre to be branded. The coefﬁcients of static
and kinetic friction are [is = 0.5 and m, = 0.4 for
all three of these critters. Bush is 8 meters East
and 6 meters North of Torino. Cheney is 12 me-
ters West and 16 meters North of Torino. Take i
to be East and to be North. A) (9 pts) What are the maximum forces, F3 and EC, magnitudes and directions,
which Bush and Cheney can each exert on Torino?
133 = usMBgmsi + 0.65) = 375(0.8€ + 0.63) Newtons = 3002 + 2253' Newtons
130 = usMcg(—0.6i+0.8§) = 500(—0.62+0.83) Newtons = ~3002+4003 Newtons B) (8 pts) What is the smallest mass which Torino can have to avoid being dragged?
35%? = 625i Newtons = usMTormogi‘ —> MW"in : (625/5)kg= 125 kg. Torino C) (9 pts) What will Torino’s acceleration, (henna, be if he has a mass of 60 kg? (Note
that if Bush and Cheney d0 manage to drag Torino their boots will not be slipping
on the dirt whereas Torino’s hooves will be slipping on the dirt.) "fm'ction : _MkMTmmog§ Newtons : —240j' Newtons. Torina ﬂow; 2 (625 — 240)} Newtons = MTaTormo —> [i : 6.42; m/sec2 D) (9 pts) How much work will Bush and Cheney EACH do if, both working together,
they manage to drag the 60 kg Torino a distance of 3 meters while maintaining their constant maximum forces? _' A
WB 2 F3 ~ 3j 2675 Joules WC 2 FC - 33' = 1200 Joules On this exam we take the acceleration of gravity at the surface of the earth to be
9 = 10 m/sec2 instead of g = 9.81 m/secz. A) (15 pts) If the radius of the earth is RE 2 6.4 X 106 meters What would the mass
of the earth be to give 9 = 10 m/sec2 at the earth’s surface? g = GMWth/Rgmh » Mm”, = 6.11 x 1024 kg. B) (15 pts) Suppose, in contrast to your answer in part “A”, that the mass of the
earth is 1.493 X 1025 kg. A marble, 3.6 X 106 meters above the surface of the earth
and with a mass of 1 gram, is initially ﬂying away from the earth at 1000 m / sec.
The earth’s gravitational attraction eventually causes the marble to collide with
the earth. What is the speed of the marble when it collides with the earth? V = 1.1 X 104 m/sec 3. Cain has a rotten grape, Mgmpe = 0.1 gram, which
he shoots horizontally at his brother, Abel, using
a rubber band which obeys Hooke’s Law. Cain
stretches the rubber band by 20 cm from its un-
stretched, equilibrium length. The time between MBDRMW
when Cain releases the grape and when it loses
contact with the rubber band is 0.05 seconds. /\\\“"m; A) (8 pts) What is the spring constant of the
T = 0.2 seconds to : 107r= Vic/m k = 9.87 x 10-2 N/m B) (9 pts) What is the speed of the grape when it leaves the rubber band? Speed : meax : 6'28 m/Sec On October 23, 1927, three days after its invention,
the first rubber band is tested. C) (9 pts) Given that Cain stretches the rubber band by 20 cm and then releases the
grape, What is the speed of the grape when the rubber band has shrunk such that
it is stretched by only half the original amount, i.e. by 10 cm? V(t) = —6.28 sin(wt) m/sec X(t) = 0.2cos(wt) meters If X(t) = 0.1 —> cos(wt) 2 0.5 —> sin(wt) : v0.7 2 0.8660
Speed 2 [V(t)[ = 5.439 m/sec. D) (9 pts) HOW much time passes from when Cain releases the grape to when the
rubber band is stretched by only 10 cm?
cos(wt) : 0.5 —) wt 2 7r/3
= 7r/(3w) = 7r/307r = 1/30 2 0.0333 seconds ...
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