Question

If F=40 N, m1=m2=20kg . The acceleration of the bocks after the force aplied is 0.5m/s^2. Calculate the frictional force . Explain Your answer.​

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

fk

=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.

Answer 2

Answer:

A block of mass M

1

=10 kg is placed on a slab of mass M

2

= 30 Kg. The slab lies on a frictionless horizontal surface as shown in figure. The coefficient of static friction between the block and slab is μ

s

= 0.25 and that of dynamic friction is μ

k

= 0.12. A force F = 40 N acts on block M

1

. The acceleration of the slab will be (g = 10 m/s

2

)