Question

A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force F⃗ as shown in the figure. The coefficient of static friction between all surfaces is 0.57 and the kinetic coefficient is 0.43. What is the minimum value of F needed to move the two blocks?

Answer 1

Answer:

$F_s = 44.7 N$

Explanation:

As we know that the force is applied of the 5 kg block

so the minimum force that is required to move the two block must be slightly more than the maximum static friction force

so here we have

$F_s = \mu_s F_n$

so we have

$F_n = (m_1 + m_2) g$

$F_n = (5 + 3)(9.8)$

$F_n = 78.4 N$

also we know that

$\mu_s = 0.57$

so we have

$F_s = (0.57)(78.4)$

$F_s = 44.7 N$

Answer 2

0.57 * (3*(3.0 kg * 9.8) + (5.0 kg * 9.8))