Question

Two blocks, stacked one on top of the other, slide on a frictionless horizontal surface. The surface between the two blocks is rough, however, with a coefficient of static friction equal to 0.50. The top block has a mass of 2.3 kg, and the bottom block's mass is 4.2 kg. If a horizontal force F is applied to the bottom block, what is the maximum value F can have before the top block begins to slip?

Answer 1

Answer:

$F_{net} = 31.88 N$

Explanation:

When top block is just or about to slide on the lower block then we can say that the frictional force on it will be maximum static friction

So we will have

$F_{net} = ma$

$F_s = ma$

$\mu_s mg = ma$

$a = \mu_s g$

$a = (0.50)(9.81)$

$a = 4.905 m/s^2$

now for the Net force on two blocks to move together

$F_{net} = (m_1 + m_2) a$

$F_{net} = (2.3 + 4.2)(4.905)$

$F_{net} = 31.88 N$