Bukhovtsev-et-al-Problems-in-Elementary-Physics

MECHANICS
205
to the water is
The total work W'=Wl-~W~+W~.
136. It is the simplest to solve the problem in a coordinate system rela..
ted to the escalator. The man will walk a distance l=_.h_+ut relative
sm a
to the escalator, where ur is the distance covered by the escalator. He should
perform the work W=(-.h_+v-r) mgsincx, since during the ascent the
sin ex
force mg is applied over the distance 1 and forms an angle of 90 0 - a with it.
Part of the work mgh is spent to increase the potential energy of the
man, and the remainder mgucsin a. together with the work of the motor
that drives the escalator is spent to overcome the forces of friction.
137. The relation between the elastic force and the deformation is shown
in Fig. 321. The work done to stretch (or compress) the spring by a small
amount 8X is shown by the area of the hatched rectangle ~W =F~x. The total
work in stretching or compressing the spring by the amount I, equal to its
potential energy Ep• is shown by the area of triangle OBC: W=Ep=k~2.
Let us recall that an expression for the distance covered in uniformly accelerated
motion can usually be obtained by similar reasoning.
138. The man acting with force F on the spring does the work WI =- FL.
At the same time, the floor of the railway carriage is acted upon from the
side of the man by the force of friction F. The work of this force W2=FL.
Therefore, the total work performed by the man in a coordinate system
related to the Earth is zero, in the same way as in a system related to the train.
139. In the system of the train the work done is equal to the potential
kl
energy of the stretched spring (see Problem 137) W= T' i
since the force of
Fig. 321
friction between the man and the floor of the railway carriage does not do
work in this system.
In the system related to the Earth,
the work the man does to stretch
F
the spring is equal to the product
of the mean force k~ and the distance
L-l, i. e., W l=k~
(L-l). The
man acts on the floor of the carriage
with the same mean force ~ . The
kl
work of this force W 2 = 2' L. The
o ~-----~~~--f---!....
IC total work in the given coordinate
klt
system W=W 1+W2 = T is the
same as in the system or the carriage.