I have a question on block that I was solving in my book. I got the answer right but different procedures for doing it I want to make sure my logic is correct,and I didn't get the answer out of probability. Two crates of mass 75kg and 110kg, are in contact and at rest on a horizontal surface. A 620N force is exerted on the 75kg crate. If the coefficient of kinetic friction is 0.15, calculate a)The acceleration of the system. Here I solved as follows:- Fnet1 = F1 - F12 - Ff1 --> F1 - f12 - Ff1 = (m1)(a); Fnet2 = F21 - Ff2 --> F21 - Ff2 = (m2)(a); We know since objects are in horizontal plane then normal force is equal to weight of objects. Fn1 = (m1)(a); --> Ff1 = (u)(m1)(a); Fn2 = (m2)(a); --> Ff2 = (u)(m2)(a); Fnet1 -> F1 - F12 - (u)(m1)(a) = (m1)(a); Fnet2 -> F21 - (u)(m2)(a) = (m2)(a); Adding both equations together we'll get net acceleration on the system. F1 - F12 - um1g + F21 - um2g = m1a + m2a; F - (u)(m1)(g) - (u)(m2)(g) = (a)(m1 + m2); (F - (u)(m1)(g) - (u)(m2)(g)) / (m1 + m2) = a; Substituting in for values we'll get the following:- a = 1.9 m/s^2 ,which is the same answer as in my book but my book got it as follows:- Fx = F1 - Fr = (m1 + m2)(a); F1 - Fr = (m1 + m2)(a) -> a = (F1 - Fr) / (m1 + m2); Substituting for values he got that answer which is the same as mine. So did I do something wrong here or maybe somewhere in my math could be modified to get that same formula. Thanks.