I'm having trouble solving for the correct answer for this problem: A block of mass M1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. It collides with block of mass M2 = 1.7 kg which is initially stationary. The blocks stick together and encounter a rough surface. The blocks eventually come to a stop after traveling a distance d = 1.85 m . What is the coefficient of kinetic friction on the rough surface? With the help the problem provided, I get that KE = (1/2)m*v^2 which also equals momentum squared over 2 times the two masses added together; this equals the post-collision KE. I also find out that the Force of friction = (friction constant)g(m1 + m2) . Then through the vector dot product you can figure out that delta KE = F*d = (sliding friction coefficient)gd(m1 + m2)cos(180) . I calculate the initial KE as 69.4575 and the post-collision KE as 46.75 leaving delta KE as 22.7073. Then when I solve for the sliding friction coefficient, from the equation delta KE = (friction coefficient)gd(m1+m2)cos(180). After my numbers are plugged in and I solve, I get 0.240859 as the answer. However, the online program I'm using says it is incorrect. Am I doing something wrong?