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NEET 2022 Model Paper 4
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© examsnet.com
Question : 9
Total: 200
Two blocks A and B of masses
2
m
and
m
respectively are connected by a massless spring of stiffness constant
k
. At the instant shown, there is force acting, on
2
m
, equal to F and acceleration of mass
m
is 'a' towards left. The accelerationof
2
m
at that instant is
2
F
+
2
m
a
3
m
F
+
2
m
a
3
m
F
−
m
a
2
m
F
+
m
a
2
m
Validate
Solution:
FBD for
m
:
f
=
m
a
...(i)
FBD for
2
m
:
F
+
f
=
2
m
×
a
′
F
+
m
a
=
2
m
×
a
′
a
′
=
F
+
m
a
2
m
© examsnet.com
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