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Calculating the kinetic energy of a package

Suppose a 30.0-kg package on the roller belt conveyor system in [link] is moving at 0.500 m/s. What is its kinetic energy?


Because the mass m and speed v are given, the kinetic energy can be calculated from its definition as given in the equation KE = 1 2 mv 2 size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } } {} .


The kinetic energy is given by

KE = 1 2 mv 2 . size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } "." } {}

Entering known values gives

KE = 0 . 5 ( 30.0 kg ) ( 0.500 m/s ) 2 , size 12{"KE"=0 "." 5 \( "30" "." 0" kg" \) \( 0 "." "500"" m/s" \) rSup { size 8{2} } ,} {}

which yields

KE = 3.75 kg m 2 /s 2 = 3.75 J. size 12{"KE"=3 "." "75"`"kg" cdot m rSup { size 8{2} } "/s" rSup { size 8{2} } =3 "." "75"`J "." } {}


Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation that people can move packages like this without exhausting themselves.

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Real world connections: center of mass

Suppose we have two experimental carts, of equal mass, latched together on a track with a compressed spring between them. When the latch is released, the spring does 10 J of work on the carts (we’ll see how in a couple of sections). The carts move relative to the spring, which is the center of mass of the system. However, the center of mass stays fixed. How can we consider the kinetic energy of this system?

By the work-energy theorem, the work done by the spring on the carts must turn into kinetic energy. So this system has 10 J of kinetic energy. The total kinetic energy of the system is the kinetic energy of the center of mass of the system relative to the fixed origin plus the kinetic energy of each cart relative to the center of mass. We know that the center of mass relative to the fixed origin does not move, and therefore all of the kinetic energy must be distributed among the carts relative to the center of mass. Since the carts have equal mass, they each receive an equal amount of kinetic energy, so each cart has 5.0 J of kinetic energy.

In our example, the forces between the spring and each cart are internal to the system. According to Newton’s third law, these internal forces will cancel since they are equal and opposite in direction. However, this does not imply that these internal forces will not do work. Thus, the change in kinetic energy of the system is caused by work done by the force of the spring, and results in the motion of the two carts relative to the center of mass.

Determining the work to accelerate a package

Suppose that you push on the 30.0-kg package in [link] with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N.

(a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force.

Strategy and Concept for (a)

This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See [link] .) As expected, the net work is the net force times distance.

Questions & Answers

Suppose a speck of dust in an electrostatic precipitator has 1.0000×1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). How many electrons does it have?
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how would I work this problem
how can you have not an integer number of protons? If, on the other hand it supposed to be 1e12, then 1.6e-19C/proton • 1e12 protons=1.6e-7 C is the charge of the protons in the speck, so the difference between this and 5e-9C is made up by electrons
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Practice Key Terms 3

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