Physics! Please help!?

[360]

Two boxes connected by a light horizontal rope are on a horizontal surface, as shown in the following figure. The coefficient of kinetic friction between each box and the surface is 0.30. One box (box B) has mass 4.76kg , and the other box (box A) has mass m . A force F with magnitude 40.1N and direction 53.1∘ above the horizontal is applied to the 4.76kg box, and both boxes move to the right with 1.50m/s2.


A.)What is the tension T in the rope that connects the boxes?
Express your answer to two significant figures and include the appropriate units.


B.)What is the mass m of the second box?
Express your answer to two significant figures and include the appropriate units.

[billrussell42]

"as shown in the following figure" ??

[Woodsman]

The setup seems clear enough.

A) B normal force Fn = mg - FsinΘ = 4.76kg * 9.8m/s² - 40.1N * sin53.1º
Fn = 14.6 N
so its friction force Fb = µ*Fn = 0.3 * 14.6N = 4.4 N

The horizontal (accelerating) component of the applied force is
Fh = 40.1N * cos53.1º = 24 N

The net force horizontally is
Fnet = ma = Fh - Fb - Fa where Fa is A's friction force.
(4.76kg + m) * 1.50m/s² = 24N - 4.4N - 0.3 * m * 9.8m/s²
Dropping units for ease (m is in kg):
7.14 + 1.5m = 19.7 - 2.94m
4.44m = 12.56
m = 2.83 kg ← (B)

tension T = Fnet + Fa = 2.83kg * 1.5m/s² + 0.3 * 2.83kg * 9.8m/s²
T = 12.6 N ← (A)

Could have found tension without finding the mass of A through the fbd for B:
Fnet = Fh - Fb - T
4.76kg * 1.5m/s² = 24N - 4.4N - T
T = 12.5 N √