Page 136 - Class-11-Physics-Part-1_Neat

122 PHYSICS

The total mechanical energies E , E , and E Answer (i) There are two external forces on
0 h H
of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the
level), h and H, are string. The latter does no work since the
displacement of the bob is always normal to the
E = mgH (6.11 a)
H string. The potential energy of the bob is thus
1
E = mgh + mv 2 (6.11 b) associated with the gravitational force only. The
h
2 h total mechanical energy E of the system is
E = (1/2) mv 2 (6.11 c)
0 f conserved. We take the potential energy of the
The constant force is a special case of a spatially system to be zero at the lowest point A. Thus,
dependent force F(x). Hence, the mechanical at A :
energy is conserved. Thus
E = E
H 0 1 2
1 2 E = mv 0 (6.12)
or, mgH = mv f 2
2
v = 2gH [Newton’s Second Law]
f
a result that was obtained in section 3.7 for a where T is the tension in the string at A. At the
A
freely falling body. highest point C, the string slackens, as the
Further, tension in the string (T ) becomes zero.
E = E C
H h Thus, at C
which implies,
2 1
2
v h = 2 g(H − h) (6.11 d) E = mv + 2mgL (6.13)
c
and is a familiar result from kinematics. 2
At the height H, the energy is purely potential. mv 2
It is partially converted to kinetic at height h and mg= L c [Newton’s Second Law] (6.14)
is fully kinetic at ground level. This illustrates
the conservation of mechanical energy. where v is the speed at C. From Eqs. (6.13) and
C
(6.14)
Example 6.7 A bob of mass m is suspended
5
t
by a light string of length L . It is imparted a E = mgL
horizontal velocity v at the lowest point A 2
o
such that it completes a semi-circular Equating this to the energy at A
trajectory in the vertical plane with the string
5 m
becoming slack only on reaching the topmost mgL = v 2 0
point, C. This is shown in Fig. 6.6. Obtain an 2 2
expression for (i) v ; (ii) the speeds at points or, v = 5gL
o 0
B and C; (iii) the ratio of the kinetic energies
(K /K ) at B and C. Comment on the nature (ii) It is clear from Eq. (6.14)
B C
of the trajectory of the bob after it reaches v C = gL
the point C.
At B, the energy is
1
E = mv + mgL
2
2 B
Equating this to the energy at A and employing
the result from (i), namely v = 5 gL ,
2
0
1 2 1 2
mv + mgL = mv
2 B 2 0
5
= m g L
Fig. 6.6 2

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