Download Assignment 4 Solutions

GSCI 1010
ASSIGNMENT 4 SOLUTIONS
08 MARCH 2010
Ted’s Questions
1. Imagine a crate being pushed across the floor of a warehouse, with four distinct forces acting on
the crate as it slides: the weight of the crate, the upward force of the floor, the horizontal
pushing force and friction. Assume the magnitudes of these forces are 980 N, 980 N, 200 N and
80 N, respectively.
(a)
Make a sketch of the crate, showing the four forces acting on it (8 marks).
+980 N
+200 N

8
–80 N
direction of motion
–980 N
(b)
Find the work done by each of the four forces as the crate travels 5 m across the floor
(8 marks). [Note: Work is given by W = Fd where F is the force and d is the distance
traveled in the direction of the force.]
4
Since the two vertical forces (980 N) are perpendicular to the motion, there is
no distance traveled in the direction of these forces. Hence, the work done by
each vertical force is zero.
2
For the +200 N pushing force (to the right), the 5 m distance traveled is in the
same direction as the force, so the work done is W = Fd = +200(5) = 1000 N-m
(or J).
2
For the –80 N friction force (to the left), the distance traveled is in the opposite
direction to the force, so the work done is W = Fd = –80(5) = –400 N-m (or J).
(Note that work is negative when the force is directed opposite to the motion.)
2
Text Problems (selected questions will be worth 19 marks)
Chapter 6:
Page 283, Question # 2
There are two key links between energy and work. First, they have the same physical
dimensions. Energy is usually specified in Joules (J), where 1 J = 1 kg-m2/s2. Work is
normally given in Newton-meters (N-m), but since F = ma, 1 N = 1 kg-m/s2. Therefore,
1 N-m = (1 kg-m/s2)-m = 1 kg-m2/s2, which is the same as 1 J.
Second, when a body gains or loses a certain amount of energy (either potential or
kinetic), this energy is exactly accounted for by the work done to bring about the change.
For example, the work done in lifting an object vertically upwards is precisely equal to the
gravitational potential energy gained as a result. Likewise, the work done by gravity as an
object falls a certain distance is equal to the resulting gain in the object’s kinetic energy.
Page 284, Problem # 4
For a typical human body mass of (say) 70 kg, the associated rest energy is
3
2
E = mc2 = 70(3.00  108)2 = 6.3  1018 J..
From Figure 6.2, this would be the order of magnitude of energy associated with a large
earthquake or hurricane.
Page 284, Question for Thought # 4
2
3
The work done by a force F is defined by W = Fd, where d is the distance traveled in the
direction of the force. For the circular orbit of a planet around its sun, the motion of planet
is always perpendicular to the sun’s gravitational pull. Hence, d = 0 and no work is done
by gravity.
If the work done by gravity were not zero, it would have to result in a change in either
gravitational potential energy or kinetic energy (as described in Question 2, above).
However, since the planet stays the same distance from the center of the sun, its
gravitational potential energy (GPE) remains constant. And, according to the satellite
equation, the speed of the planet in its orbit, and hence its kinetic energy (KE) is also
constant. The fact that both GPE and KE are unchanged is therefore consistent with the
fact that the sun’s gravity does no work.
3
Chapter 17:
Page 656, Question # 2
1
1
1
1
It is through advances in high-energy physics that we have come to discover the existence
of leptons and quarks. In turn, this has helped us understand how the very early universe
evolved, including: (i) the appearance of electrons, neutrinos and their anti-particles (at
age 10-6 seconds); (ii) the creation, from quarks, of all of the universe’s protons and
neutrons (at age 2 seconds); (iii) the formation, through “fusion”, of the first atomic nuclei
(deuterium and helium, at age 3 minutes); and finally, (iv) the capture of electrons by these
nuclei to form “conventional matter” in the form of atoms of hydrogen and helium (the socalled “dawn of light”, at age 106 years).
Page 656, Question # 8
2
1
2
White dwarfs, neutron stars and black holes are similar in that they are all final “death”
states of stars which have come to the end of their evolutionary cycles. Specifically, at the
end of the “red giant” stage, when fusion of carbon into iron begins to dominate in the
core, all stars commence their final “collapse”. It is at this point that the differences
between these end states will become apparent. Small stars, like our own Sun, will
collapse slowly and gently to form small but extremely hot white dwarfs. Much larger stars
will collapse very rapidly and experience a catastrophic supernova explosion which blows
most of the star outward, with only a small part of the core continuing to collapse. Smaller
collapsing core remnants will stop when the atoms have been crushed into a solid ball of
neutrons (i.e. a neutron star). For larger cores, the neutron ball itself is crushed, with the
resulting final object becoming so tiny and so dense that light can no longer escape, which
is why it is called a black hole.