Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
A particle acted upon by constant forces ( 4 hati+ hatj-3 hatk ) and ( 3 hati+ hatj- hatk ) is displaced from the point ( hati+2 hatj+3 hatk ) to the, point (5 hati+4 hatj+ hatk ) . The total work done by the forces in SI unit is
Q. A particle acted upon by constant forces
(
4
i
^
+
j
^
−
3
k
^
)
and
(
3
i
^
+
j
^
−
k
^
)
is displaced from the point
(
i
^
+
2
j
^
+
3
k
^
)
to the, point
(
5
i
^
+
4
j
^
+
k
^
)
. The total work done by the forces in
S
I
unit is
4312
267
KEAM
KEAM 2008
Work, Energy and Power
Report Error
A
20
18%
B
40
46%
C
50
8%
D
30
13%
E
35
13%
Solution:
Here,
F
1
=
4
i
^
+
j
^
−
3
k
^
,
F
2
=
3
i
^
+
j
^
−
k
^
r
1
=
i
^
+
2
j
^
+
3
k
^
,
r
2
=
5
i
^
+
4
j
^
+
k
^
Displacement,
r
=
r
2
−
r
1
=
(
5
i
^
+
5
j
^
+
k
^
)
−
(
i
^
+
2
j
^
+
3
k
^
)
=
4
i
^
+
2
j
^
−
2
k
^
Work done by the forces,
W
=
[
F
1
+
F
2
]
⋅
r
=
[
(
4
i
^
+
j
^
−
3
k
^
)
(
3
i
^
+
j
^
−
k
^
)
]
⋅
(
4
i
^
+
2
j
^
−
2
k
^
)
=
(
7
i
^
+
2
j
^
−
4
k
^
)
⋅
(
4
i
^
+
2
j
^
−
2
k
^
)
=
28
+
4
+
8
=
40
J