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Question

The potential energy of a body of mass 2 kg in a conservative field is U=(6x8y) J. If the initial velocity of the body is u=(1.5^i+2^j) m/s, then the total distance travelled by the particle in the first two seconds is

A
10 m
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B
12 m
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C
15 m
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D
18 m
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Solution

The correct option is C 15 m
From relation
F=δUδx^iδUδy^j
F=(6^i+8^j) N
a=Fm=6^i+8^j2=(3^i+4^j) m/s2

Initial velocity, u=32^i+2^j
u=3^i+4^j2m/s

Hence, vector u anda are in the same direction. Therefore the body will accelerate along a straight line path
( angle between a and u is θ=0)

|a|=32+42=5 m/s2=constant
& |u|=52=2.5 m/s

Applying kinematic equation, the distance travelled by the body in the first 2 sec will be:
s=ut+12at2
s=(2.5×2)+12×5×4
s=15 m

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