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AIEEE 2006 Solved Paper
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© examsnet.com
Question : 62
Total: 143
The potential energy of a
1
kg
particle free to move along the
x
-axis is given by
V
(
x
)
=
(
x
4
4
−
x
2
2
)
J
The total mechanical energy of the particle is
2
J
. Then, the maximum speed (in
m
∕
s
) is
3
√
2
√
2
1
√
2
2
Validate
Solution:
Velocity is maximum when kinetic energy is maximum and when kinetic energy is maximum then potential energy should be minimum
For minimum potential energy,
For minimum potential energy,
d
V
d
x
=
0
⇒
x
3
−
x
=
0
⇒
x
=
±
1
⇒ Min. Potential energy (P.E.)
;
=
1
4
−
1
2
=
−
1
4
J
K
⋅
E
⋅
(
max
.
)
+
P
⋅
E
⋅
(
min
.
)
=
2
𝜕(Given)
∴
K
⋅
E
⋅
(
max
.
)
=
2
+
1
4
=
9
4
∴
1
2
m
v
max
2
=
9
4
⇒
1
2
×
1
×
v
2
max
.
=
9
4
⇒
v
max
=
3
√
2
m
/
s
© examsnet.com
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