## Essential University Physics: Volume 1 (3rd Edition)

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 $a=\frac{F_{net}}{m}=\frac{2.224N}{2.27Kg}=.99\frac{m}{s^2}$