Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
The potential energy for a force field vecF is given by U(x, y) = cos (x+ y). The force acting on a particle at position given by coordinates (0, π / 4) is -
Q. The potential energy for a force field
F
is given by
U
(
x
,
y
)
=
cos
(
x
+
y
)
. The force acting on a particle at position given by coordinates
(
0
,
π
/4
)
is -
5277
240
BITSAT
BITSAT 2010
Report Error
A
−
2
1
(
i
^
+
j
^
)
21%
B
2
1
(
i
^
+
j
^
)
63%
C
(
2
1
i
^
+
2
3
j
^
)
11%
D
(
2
1
i
^
−
2
3
j
^
)
5%
Solution:
F
x
=
−
∂
x
∂
U
=
sin
(
x
+
y
)
F
y
=
−
∂
x
∂
U
=
sin
(
x
+
y
)
F
x
=
sin
(
x
+
y
)
]
(
0
,
π
/4
)
=
2
1
,
F
y
=
sin
(
x
+
y
)
]
(
0
,
π
/4
)
=
2
1
∴
F
=
2
1
(
i
^
+
j
^
)