The work done on a particle
of mass m by a force
u
tenal
K
У
chou
no di
+y?) (x+y)
(K
being the constant of
appropriate dimensions,
when the particle is taken
from the point (a, 0) to the
point (0, a) along a circular
path of radius a about the
origin in the x - y plane is
Answers
Step-by-step explanation:
Given that,
Force F=K
⎣
⎢
⎢
⎢
⎡
(x
2
+y
2
)
2
3
x
i
^
+
(x
2
+y
2
)
2
3
y
j
⎦
⎥
⎥
⎥
⎤
Now, for the small distance is dr travelled by the particle in the direction
d
r
=d
x
i
^
+d
y
j
^
Now, small work done is
dw=F.d
r
The total work done
W=∫dw
W=∫F⋅d
r
W=∫K
⎣
⎢
⎢
⎢
⎡
(x
2
+y
2
)
2
3
x
i
^
+
(x
2
+y
2
)
2
3
y
j
⎦
⎥
⎥
⎥
⎤
⋅[dx
i
^
+dy
j
^
]
W=K
a
∫
0
⎣
⎢
⎢
⎢
⎡
(x
2
+y
2
)
2
3
x
i
^
⎦
⎥
⎥
⎥
⎤
+
0
∫
a
⎣
⎢
⎢
⎢
⎡
(x
2
+y
2
)
2
3
y
j
⎦
⎥
⎥
⎥
⎤
Now, solve the first term of integration
Put,
x
2
+y
2
=t
2xdx=dt
Now,
=
2
K
a
∫
0
(x
2
+y
2
)
2
3
2xdx
=
2
K
a
∫
0
(t)
2
3
dt
=
2
K
×2a
2
1
=Ka
2
1
Now, similarly for second term of integration
=−Ka
2
1
Now, work done is
W=Ka
2
1
−Ka
2
1
W=0
Hence, the work done on a particle is 0