Question

The lower block in the figure is pulled on by a rope with a tension force of 20 N. The coefficient of kinetic friction between the lower block and the surface is 0.35. The coefficient of kinetic friction between the lower block and the upper block is also 0.35. A. What is the acceleration of the 2.0 kg block?

Answer 1

Equation for block 1 is F = M*A The force for block 1 is Tension - (friction coefficent*mass*g) the mass is 1 kg the a is unknown, but we know that both blocks have the same acceleration, since they are connected. So here is the final equation that describes the acceleration of block 1: T-.3(1kg*9.8) =1kg*a T-.3(9.8) = a (I just got rid of the 1kg, because 1 * anything is itself.) now lets find an equation that describes the acceleration of block 2: F = ma 20N - T -(.3*3kg*9.8 + .3(1kg*9.8) = 2kg*a so: a = {20N - T -(.3*3kg*9.8 + .3(1kg*9.8) } / 2 since a is the same for block a and b, set these two equations equal to each other, and solve for T. this gives T = 4.706. plug T back in one of the equations, and get a. The first equation is easier to work with: T - (.3*g) = a 4.706 -(2.94) = a a = 1.766