Question

Two block of masses m1 = 3 kg and m2 = 4 kg are touching each other on a frictionless table, as shown in figure. If the force shown acting on m1 is 5 N (a) What is the acceleration of the two blocks and (b) What is the force exerted by m1 on m2 ?

Answer 1

Since the floor is given friction less So we can use Newton's II law to find the acceleration of the system

$F = (m_1 + m_2)a$

$5 = (4 + 3)a$

$a = \frac{5}{7} m/s^2$

So the acceleration of both blocks will be 5/7 m/s^2

now in order to find the force that mass m1 exert on mass m2 is given as

$F_n = m_2*a$

$F_n = 4*\frac{5}{7}$

$F_n = \frac{20}{7} N$

so the normal force due to m1 on block m2 is given as 20/7 N