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

176 ANSWERS AND SOLUTIONS
ice
dry snow
wet wood block
dry wood block
wet asphalt
dry asphalt
dry concrete
k=O.l
k=0.2
k=O.3
k=O.5
k=O.4
k=O.6
k=O.7
The coefficient of friction does not depend on the speed if the accuracy to
the first digit after the decimal point is wanted.
49. When the motor vehicle accelerates, the rear wall of the fuel tank
imparts an acceleration vlt to the petrol. According to Newton's second law,
the force F required for this acceleration is Alp ~ J where A is the area of
the rear wall of the fuel tank. In conformity with Newton's third law, the
petrol will act with the same force on the wall. The hydrostatic pressure of
the petrol on both walls is the same. Hence, the difference of pressures exerted
on the walls is Ap= ~ = Ip -7-.
M
50. The mass of the left..hand part of the rod mt=T I and of the
right-hand part m2= ~ (L-I), where M is the mass ()f the entire rod.
Under the action of the appl ied forces each part of the rod moves with the
same acceleration a. Therefore,
Ft-F=mta
F-F 1 = m2a
Hence the force F is
B
Fig. 284
F
51. The motion of the ball will be uniform. The images of the ball on the
film appear at intervals of t = 1/24 s.
The distance between the positions A and B of the ball in space that
correspond to the positions C and D of the images on the film is AB=CD ~; •
as shown in Fig. 284. The focal length of the lens OF= 10 em, OE= 15 metres,
and CD=3 rnrn, The velocity of the ball
A
AB
vt = - t- = 10.8 m/s.
When the ball is in uniform motion. mg=
=Iro~. In the second case 4 mg=kv:. Hence,
E v:=4v~ and v 2=21.6 tnls.
G
k=- ~ 3.9XIO- o kgf·s 2/m2
v~
52. Figure 285 shows the forces that act
on the weights. The equations of motion for