Question

two blocks of masses 3kg and 2kg are paced beside each other in contact with each other on a rough horizontal surface .a horizontal force of 20N is applier on 2kg the coefficient of friction is 0.1and g=10 the force of contact between two blocks

Answer 1

Total friction force acting on two blocks will be

$F_f = \mu*m_1g + \mu*m_2g$

$F_f = 0.1*(2+3)*10 = 5 N$

now using II law of Newton

$F_{net} = F - F_f$

$F_{net} = 20 - 5 = 15 N$

as we know that

$F_{net} = m*a$

$15 = (2+3)*a$

$a = 3 m/s^2$

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

$F_{net} = m*a$

$F_n - F_f = m*a$

$F_n - 0.1*3*10 = 3* 3$

$F_n = 3 + 9 = 12 N$

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

Answer 2

Answer:

8 newtons

Explanation: