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JEE Advanced 2020 Concept Recapitulation Test 4 Paper 1
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© examsnet.com
Question : 8
Total: 54
One end of a spring of force constant
25
N
∕
m
is connected with a block of mass
1
k
g
and the other end is fixed to a vertical wall. The block is given velocity
6
m
∕
s
towards the wall (when the spring is in natural length) on a horizontal rough surface having coefficient of friction
0.55
. Then:
The compression in the spring when the block stops momentarily for the
l
st
time is
1
m
.
The compression in the spring when the block stops momentarily for the
I
st
time is
0.5
m
.
The acceleration of the block when it stops momentarily for the
I
n
d
time is
12.5
m
∕
s
2
The acceleration of the block when it stops momentarily for the
I
n
d
time is
8.5
m
∕
s
2
Validate
Solution:
Using work energy theorem
−
µ
m
g
x
−
1
2
k
x
2
=
−
1
2
m
v
2
11
x
+
25
x
2
−
36
=
0
⇒
x
=
1
m
Now when it stops momentary for the second time
−
µ
m
g
d
−
[
1
2
k
d
2
2
−
1
2
k
d
1
2
]
=
0
11
d
+
25
d
2
2
−
25
d
1
2
=
0
11
d
1
+
11
d
2
+
25
d
2
2
−
25
d
1
2
=
0
So,
25
d
2
2
+
11
d
2
−
14
=
0
[
as
d
=
1
m
]
d
2
=
0.56
m
So acceleration
a
=
k
d
2
−
µ
m
g
m
a
=
8.5
m
∕
s
2
© examsnet.com
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