Question

Block A, mass 250 g, sits on top of block B, mass 2.0 kg. The coefficients of static and kinetic friction between blocks A and B are 0.34 and 0.23, respectively. Block B sits on a frictionless surface. What is the maximum horizontal force that can be applied to block B, without block A slipping?

Answer 1

mass of block A = 250 g = 0.25 kg

mass of block B = 2 kg

coefficient of kinetic friction = 0.23

coefficient of static friction = 0.34

now, if block is not supposed to slip , then force experienced by A must be less than coefficient of static friction times the weight

= 0.34 * 0.25*9.8

= 0.833 N

therefore, a = 0.833/0.25

a = 3.332 m/s^2

Now, F_applied = (m1+m2) a

F_applied = (2+0.25)*3.332

F_applied = 7.5 N (Maximum horizontal force that can be applied to block B)

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