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)

Please ask your doubts or queries in the comment section below.

Please kindly upvote if you are satisfied with the solution.

Thank you.

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