#### Answer

Please see the work below.

#### Work Step by Step

We know that the net force during the interval B is
$F_{net}=W_{apparent}-W_{actual}$
We plug in the known value to obtain:
$F_{net}=5.5-5=0.5lb=(0.5lb)(\frac{4.448N}{1lb})=2.224N$
The mass of the laptop is given as
$m=5lb=(5lb)(\frac{0.454Kg}{1lb})=2.27Kg$
The acceleration is given as
$a=\frac{F_{net}}{m}=\frac{2.224N}{2.27Kg}=.99\frac{m}{s^2}$