Question

A block of mass m = 2 kg is held in contact with a block of mass M = 10 kg by applying a horizontal force F on it Block M is lying on a horizontal frictionless surface. The coefficient of friction between the blocks is µ = 0.4. Find the minimum value of F required to hold m against M.

Answer 1

Answer:

Static Friction between the block is μ

k

M

1

g=.25×100=25N

static friction < applied force

so in this case kinetic friction will be act

f

k

=.12×100N=12N

FBD diagram of both the block as shown

for M

1

block which as acceleration a

1

F−f

K

=10a

1

eq(1)..

40−12=10a

1

a_{1}=2.8ms^{-2}

for M

2

slab which as acceleration

f

k

=30a

2

eq(2)..

12=30a

2

a

2

=0.4ms

−2

solving eq(1)and eq(2)

accleration of slab is 0.4ms

−2

Hence the B option is correct.