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Standard XII
Maths
Question
A particle moves from a point
(
2
^
i
+
5
^
j
)
to
(
4
^
j
+
3
^
k
)
when a force of
(
4
^
i
+
3
^
j
)
N
is applied. How much work has been done by the force?
-8J
-11J
2J
5J
A
2J
B
-8J
C
-11J
D
5J
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Solution
Verified by Toppr
Displacement of the particle
→
S
=
(
4
^
j
+
3
^
k
)
−
(
2
^
i
+
5
^
j
)
=
−
2
^
i
−
^
j
+
3
^
k
Force acting on the particle
→
F
=
4
^
i
+
3
^
j
∴
Work done
W
=
→
F
.
→
S
=
(
4
^
i
+
3
^
j
)
.
(
−
2
^
i
−
^
j
+
3
^
k
)
=
−
8
+
(
−
3
)
=
−
11
J
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