AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Condition for the minimum potential energy is,
Umin=‌

dU(x)

dx

0=‌

d

dx

[‌

x2

2

−x]
0=x−1
x=1
The minimum potential energy is calculated as
Umin‌‌=‌

(1)2

2

−1
‌‌=−‌

1

2

The expression of the kinetic energy is
KE‌‌=‌

1

2

mv2
‌‌=‌

1

2

(1)v2
‌‌=‌

v2

2

The total mechanical energy of the system is given by,
E=U+KE
Substitute the given values in the above equation.
E‌‌=U+KE
2‌‌=−‌

1

2

+‌

v2

2

v‌‌=√5m∕s