Question

Two blocks of masses 3 kg and 2 kg are placed beside each other contact with each other on a rough horizontal surface.  A horizontal force of 20N is applied on 3 kg.  the coefficient of friction between blocks and the surface is 0.1 and .  The force of contact between the two blocks is​

Answer 1

Answer:

Total friction force acting on two blocks will be

F_f = \mu*m_1g + \mu*m_2gF

f

=μ∗m

1

g+μ∗m

2

g

F_f = 0.1*(2+3)*10 = 5 NF

f

=0.1∗(2+3)∗10=5N

now using II law of Newton

F_{net} = F - F_fF

net

=F−F

f

F_{net} = 20 - 5 = 15 NF

net

=20−5=15N

as we know that

F_{net} = m*aF

net

=m∗a

15 = (2+3)*a15=(2+3)∗a

a = 3 m/s^2a=3m/s

2

now using the force diagram of 3 kg block we can say

F_{net} = m*aF

net

=m∗a

F_n - F_f = m*aF

n

−F

f

=m∗a

F_n - 0.1*3*10 = 3* 3F

n

−0.1∗3∗10=3∗3

F_n = 3 + 9 = 12 NF

n

=3+9=12N

so the normal force or contact force between two blocks will be 12 N