Solving a Physics Problem Involving Two Blocks Connected by a Cord

[Edwardo_Elric]

Homework Statement

A block with mass 6.00kg resting on a horizontal surface is connected by a horizontal cord passing over a light frictionless pulley to a hanging block with mass 4.00kg. The coefficient of kinetic friction between the block and the horizontal surface is 0.50. After the blocks are released,
a.) Find the acceleration of each block
b.) The tension in the cord.

Homework Equations

The Attempt at a Solution

a.)
uk = 0.50
fk = ukN
fk = 0.50(6.00kg)(9.8m/s^2)
Summation of the mass of 6.00kg block
m1a = T - fk
m1a = T - 0.50(6.00kg)(9.8m/s^2)

Summation of the mass of 4kg block
m2a = m2g - T

so i add both the two forces:
m1a = T - 0.50(6.00kg)(9.8m/s^2)
+m2a = m2g - T
________________
(m1+m2)a = m2g - 0.50(6.00kg)(9.8m/s^2)
a = { (4.00kg)(9.8m/s^2) - 0.50(6.00kg)(9.8m/s^2) } / {6.00kg+4.00kg}
a = 0.98m/s^2

dont know how the back of the book got an answer of 0.1m/s^2

b.) m2a = m2g - T
(4.00kg)(.98m/s^2) = (4.00kg)(9.8) - T
T = 35.28N

 

[Doc Al]

Your method and answers look correct to me. (Must be a typo in the book.)