Physics Problem Solutions: Work, Energy, and Momentum, Assignments of Classical Physics

Solutions to various physics problems involving work, energy, and momentum. Topics covered include the relationship between work and energy, conservation of momentum, and circular motion. Students are encouraged to use approximations and understand the logical problem-solving process.

Typology: Assignments

Pre 2010

Uploaded on 07/30/2009

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HW2 Solutions
E1.34 You are doing work on the saw (positive work) both when pushing and pulling, since
the displacement of the saw is always in the same direction as the force that you are
applying.
E2.16 The opener pivots about the end resting on the bottle cap. The hook under the cap
moves a small distance and exerts a large force on the edge of the cap, while the other
end moves a longer distance and the user’s hand can apply a correspondingly smaller
force. In other words the two torques about the pivot point must be equal.
E2.22 The horse’s energy is converted into heat (thermal energy) by the friction in the wheel
hubs.
E2.26 The road surface exerts a horizontal force of static friction on the bicycle tires, pushing
them forward.
(There are some subtle aspects of this question and the question of how energy is
supplied to the system of you and the bike. You might want to search online to see what
is discussed about this situation. The friction force from the road does not do any work
on the bicycle (can you see why?) so the energy comes from the chemical energy of
your body, transferred by your muscles into the rotational and translational kinetic
energy of the bike and the kinetic energy of your legs moving up and down.)
E2.28 Since the sled is not accelerating the net force on it must be zero. Therefore, the force
that you exert on the sled must be equal and opposite to the force of sliding friction on
its runners.
P1.18 a) The gravitational potential energy is mgh = (1000 kg)(10 m/s2)(200 m) = 2 x 106
joules.
(I have approximated g = 9.8 m/s2 by 10 m/s2. Reasonable approximations of this kind
are always acceptable in this course for numerical problems. The important point is to
understand how to solve a problem in a logical manner and to get numbers that are
good approximations for the numerical answer.)
b) From the energy table we have that a 1g cookie has an energy content of 21,000
joules. So a 5 g cookie will have an energy content of 105,000 joules. Therefore, the
number of cookies required to match the energy of part (a) will be 2,000,000/105,000
or approximately 20 cookies. This is a rather remarkable result, don’t you think? Now
you can see why eating too many chocolate chip cookies will increase your body size
significantly unless you burn up the food calories with some exercise!!
P2.13 Conservation of momentum means that your friend’s momentum must be 450 kg m/s to
the Right. The system consisting of you and your friend had zero total momentum before
you pushed on one another and this cannot change unless an external force is applied to
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HW2 Solutions

E1.34 You are doing work on the saw (positive work) both when pushing and pulling, since the displacement of the saw is always in the same direction as the force that you are applying.

E2.16 The opener pivots about the end resting on the bottle cap. The hook under the cap moves a small distance and exerts a large force on the edge of the cap, while the other end moves a longer distance and the user’s hand can apply a correspondingly smaller force. In other words the two torques about the pivot point must be equal.

E2.22 The horse’s energy is converted into heat (thermal energy) by the friction in the wheel hubs.

E2.26 The road surface exerts a horizontal force of static friction on the bicycle tires, pushing them forward.

(There are some subtle aspects of this question and the question of how energy is supplied to the system of you and the bike. You might want to search online to see what is discussed about this situation. The friction force from the road does not do any work on the bicycle (can you see why?) so the energy comes from the chemical energy of your body, transferred by your muscles into the rotational and translational kinetic energy of the bike and the kinetic energy of your legs moving up and down.)

E2.28 Since the sled is not accelerating the net force on it must be zero. Therefore, the force that you exert on the sled must be equal and opposite to the force of sliding friction on its runners.

P1.18 a) The gravitational potential energy is mgh = (1000 kg)(10 m/s^2 )(200 m) = 2 x 10^6 joules.

(I have approximated g = 9.8 m/s^2 by 10 m/s^2. Reasonable approximations of this kind are always acceptable in this course for numerical problems. The important point is to understand how to solve a problem in a logical manner and to get numbers that are good approximations for the numerical answer.)

b) From the energy table we have that a 1g cookie has an energy content of 21, joules. So a 5 g cookie will have an energy content of 105,000 joules. Therefore, the number of cookies required to match the energy of part (a) will be 2,000,000/105, or approximately 20 cookies. This is a rather remarkable result, don’t you think? Now you can see why eating too many chocolate chip cookies will increase your body size significantly unless you burn up the food calories with some exercise!!

P2.13 Conservation of momentum means that your friend’s momentum must be 450 kg m/s to the Right. The system consisting of you and your friend had zero total momentum before you pushed on one another and this cannot change unless an external force is applied to

the system. The forces you applied to each other are internal to the system and cannot change the total momentum of the system. Therefore, in this case any momentum you gain from internal forces must be equal in magnitude and opposite in direction to the momentum gained by your friend.

P3.6 a) For circular motion, the centripetal acceleration of the satellite in this example (g) is given by (speed)^2 / earth radius (R) See equation 3.3.1 in the text. So,

(speed)^2 = (R) x (g) = (6400 x 10^3 m) x (10 m/s^2 ) = 64 x 10^6 m^2 /s^2. Taking the square root gives approximately: speed (V) = 8000 m/s

b) The distance around the earth is the circumference of a circle of radius R = 2πR or 2R x 3.14. So the time for the satellite to circle the earth once is (2R x 3.14)/V. Putting in the numbers gives about 5000 sec or 83 minutes.

C1.5 a) By Newton’s 3rd^ law, the air exerts an equal and opposite force on the plane, i.e., 250,000 N in the forward direction.

b) In order to produce the takeoff energy of 5 x 10^7 joules (J), the force must do this amount of work. Since Work = (Force) x (displacement along the force), the displacement is Work/Force = 5 x 10^7 J/2.5 x 10^5 N =200 meters.

c) The total force must be (Work)/(displacement) = 5 x 10^7 J/100 m =500,000 N, so the catapult must exert 250,000 N.

d) The plane moves in the same direction as the force exerted by the catapult, so the catapult does positive work on the plane, and hence transfers positive energy to it.

e) On landing, the catapult moves in the same direction as the force the plane exerts on it, so the plane does positive work on the catapult, transferring energy to it.