Manipal Entrance Test 2015 Physics Paper

Maximum frictional force between the slab and the block
fmax=µN=µmg =

1

4

×4×10 =10N
Evidently, f∝fmax
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=Ma=

2

3

×5=

10

3

N
Using S=

1

2

at2
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

=5.55J