Answer
(a) $0.90m$
(b) $0.13m$
Work Step by Step
(a) We know that
$W=F(x_f-x_i)$
We plug in the known values to obtain:
$0.21J=(0.8N)(0.50m-0.4m)+(0.4N)(0.75m-0.5m)+(0.2N)(d-0.75m)$
$\implies 0.21J=0.03J+(0.2N)d$
$\implies d=\frac{0.21J-0.03J}{0.2N}$
$d=0.90m$
(a) We know that
$W=F(x_f-x_i)$
We plug in the known values to obtain:
$-0.19J=(0.8N)(-0.25m-0.40m)+(0.6N)(d-0.25m)$
$(0.6N)d=0.12J+0.15J-0.19J$
$d=\frac{0.08J}{0.6N}$
$d=0.13m$
