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Manipal Entrance Test 2015 Physics Paper
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© examsnet.com
Question : 36
Total: 50
A large slab of mass
5
kg
lies on a smooth horizontal surface, with a block of mass
4
kg
lying on the top of it, the coefficient of friction between the block and the slab is If the block is pulled horizontally by force of
F
=
6
N
,
the work done by the force of friction on the slab between the instants
t
=
2
s
and
t
=
3
s
is
2.4 J
5.55 J
4.44 J
10 J
Validate
Solution:
Maximum frictional force between the slab and the block
f
max
=
µ
N
=
µ
m
g
=
1
4
×
4
×
10
=
10
N
Evidently,
f
∝
f
max
So, the two bodies will move together as a single unit. If a be their combined acceleration, then
a
=
F
m
+
M
=
6
4
+
5
=
2
3
ms
−
2
Therefore, frictional force acting can be obtained as
f
=
M
a
=
2
3
×
5
=
10
3
N
Using
S
=
1
2
a
t
2
S
(
2
)
=
1
2
×
2
3
(
2
)
2
=
4
3
and
S
(
3
)
=
1
2
×
2
3
(
3
)
2
=
3
Therefore, work done by friction
=
F
[
S
(
3
)
−
S
(
2
)
]
=
10
3
[
3
−
4
3
]
=
50
9
=
5.55
J
© examsnet.com
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