MHB The work energy principle and power

[Shah 72]

A box of mass 25kg is pulled 5m across a smooth floor by rope with tension 22N. The rope is inclined at 40 degree to above the horizontal. There is a frictional force with average value 12N. The box starts from rest. Find the final speed.
Iam getting the ans 1.97m/s.
The textbook ans is 1.39 m/s
Pls help

 

[skeeter]

$W_{net} = [22\cos(40) - 12]\cdot 5 = \dfrac{1}{2} \cdot 25 \cdot v_f^2 \implies v_f = 1.39 \, m/s$

 

[Shah 72]

$W_{net} = [22\cos(40) - 12]\cdot 5 = \dfrac{1}{2} \cdot 25 \cdot v_f^2 \implies v_f = 1.39 \, m/s$

Thank you!