Answered By Majid B

 $F_g=$‍F_g= Real Weight

 $F_g'=$F_g'= Apparent Weight

Positive Direction: Downward

$F_g=mg$Fg=mg  =>   $m=\frac{F_g}{g}=\frac{750\left(N\right)}{9.81\left(\frac{m}{s^2}\right)}=76.45\left(kg\right)$‍m=F_gg =750(N)9.81(ms² ) =76.45(kg) 

$F_{net}=ma$‍F_net=ma    =>    $F_g-F_g'=ma$F_g−F_g'=ma    =>     $a=\frac{\left(F_g-F_g'\right)}{m}=\frac{\left(750N-672N\right)}{76.45kg}=1.02\left(\frac{m}{s^2}\right)$a=(F_g−F_g')m =(750N−672N)76.45kg =1.02(ms² ) 

Answered By Dhruvin P

w_apparent = w_real - ma.

First, we need to figure out the mass of the passanger. 

The passangers weight is 750 N when the speed is at a constant speed, and take the value of g= 9.81 m/s²

^{750 = (m)(9.81) => m = 76.45 kg} 

Substituing the values,

 672 N = 750 N - (76.45)(a) 

 $\frac{\left(672-750\right)}{-76.45}=1.02\left(\frac{m}{s^2}\right)$(672−750)−76.45 =1.02(ms² )