Preparation problems
5-19. My spaceship crashes on one of the Sun's nine
planets. Fortunately, the ship's scales are intact, and show that my
weight is 532 N. If I know my mass to be 60 kg, where am I
(Hint: Consult Appendix E)
Solution:
This problem is a question about the difference between weight
and mass. Looking at our constants and equation sheet, we see that near
the surface of Earth (or any other planet)
where W represents our weight, while g is the
acceleration due to gravity near the surface of the planet. Since we
know both W and m, we can quickly solve for g.
Recalling that both W and g point downwards, we
can drop the vector signs and solve for the magnitude, g.
Looking in Appendix E, we see that the spaceship is stranded on
Venus.
5-20. If I can barely lift a 50-kg concrete block on
Earth, how massive a block can I lift on the moon?
Solution:
This problem is a question about weight and mass. Looking at
our constants and equation sheet, we see that near the surface of Earth
(or any other planet)
where W represents our
weight, while g is the acceleration due to gravity near the
surface of the planet. We know that what we can lift depends only on the
amount of force we can apply to an object, which would correspond to its
weight. So if we can find the weight of the 50 kg block, we can find out
the corresponding amount of mass on the moon. Since we know both
g and m, we can quickly solve for W. Recalling
that both W and g point downwards, we can drop the
vector signs and solve for the magnitude, W.
So now we can find the mass which has a weight of 490 N on the moon. We
can look up the value of gMoon in Appendix E, which tells us
gMoon = 1.62 m/s2. To solve the problem, we use the same equation
5-22. I weigh 640 N. What's my mass?
Solution:
This problem is a question about weight and mass. Looking at
our constants and equation sheet, we see that near the surface of Earth
(or any other planet)
where W represents our weight, while g is the
acceleration due to gravity near the surface of Earth. Remembering that
both W and g point downwards, we can drop the
vector symbols, and we can then quickly solve for the mass
5-50. A spring with spring constant k = 340 N/m is used
to weigh a 6.7-kg fish. How far does the spring stretch?
Solution:
Again we have to deal with the difference between weight and mass. Looking at
our constants and equation sheet, we see that near the surface of Earth
(or any other planet)
where W represents our weight, while g is the
acceleration due to gravity near the surface of Earth. Now the spring
used in the scale will have a force applied to it exactly equal to
W. The spring will then have to exert a force equal to, but
opposite in direction to, the weight of the fish. We can then solve for
the distance the spring stretches using Hooke's Law, from our constants
and equations sheet,
We must remember that the weight of the fish will point downwards,
while the force that the spring applies points upwards. If we call up
the direction of positive displacement, we see that a positive force
exerted by the spring results in the spring being stretched downwards,
as we might expect.
Now we can put the two equations together, and solve for x:
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(6.7 kg) (9.8 m/s2)
340 N/m
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Solutions translated from TEX by TTH, version 1.57.
