Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

Solutions 131

of conservation of mechanical energy). The fre-

quency of the corresponding vibratory motion can

be found from the relation kl m = gb cos a =---

Therefore, having covered the distance xo =

(1/b) tan a after passing the "equilibrium posi-

tion", the body comes to rest. At this moment, the

restoring force "vanishes" since it is just the force

of sliding friction. As soon as the body stops, slid-

ing friction changes the direction and becomes

static friction equal to mg sin a. The coefficient of

friction between the body and the inclined plane

at the point where the body stops is Rst, = b • 2x 0 =

2 tan a, i.e. is more than enough for the body to

remain at rest.

Using the vibrational approach to the descrip-

tion of this motion, we find that the total time of

motion of the body is equal to half the "period of

vibrations". - Therefore,

T 2n

I=

2 = loro = V gb cos a •

1.30. The friction Fr,. (x) of the loaded sledge is

directly proportional to the length x of the part of

the sledge stuck in the sand. We write the equation

of motion for the sledge decelerated in the sand

in the first case:

X

ma = — mg

where m is the mass, a the acceleration, 1 the

length, andμ the coefficient of friction of the

sledge against the sand. As in the solution of

Problem 1.29, we obtained an "equation of vibra-

tions". Therefore, the deceleration of the sledge

stuck in the sand corresponds to the motion of 'a

load on a spring (of rigidity k=(mg11) p,) having

acquired the velocity vo in the equilibrium position.

Then the time dependence x (t) of the part of the

sledge stuck in the sand and its velocity v (t) can

be written as

x (t) = xo sin coot, v (t) = vo cos 6.) 0 t,

9*