Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 135: 21

$1.23\,m/s^{2}$

According to Newton's second law, $F_{net}=ma$ $\implies a=\frac{F_{net}}{m}$ $F_{net}=900\,N-800\,N=100\,N$ $m=\frac{W}{g}=\frac{800\,N}{9.80\,m/s^{2}}=81.6\,kg$ Therefore, $a=\frac{100\,N}{81.6\,kg}=1.23\,m/s^{2}$