Answer
(a) $0.45J$
(b) $0.24J$
Work Step by Step
(a) We can find the required work done as follows:
$W_1=F_1d_1$
We plug in the known values to obtain:
$W_1=(0.6N)(0.25-0m)=0.15J$
$W_2=F_2d_2$
We plug in the known values to obtain:
$W_2=(0.4N)(0.5-0.25m)=0.1J$
$W_3=F_3d_3$
We plug in the known values to obtain:
$W_3=(0.8N)(0.75-0.5)=0.2J$
Now the net work done is given as
$W=W_1+W_2+W_3$
We plug in the known values to obtain:
$W=0.15+0.1+0.2=0.45J$
(b) We know that
$W_1=F_1d_1$
$W_1=(0.6N)(0.25m-0.15m)=0.06J$
$W_2=0.1J$
and $W_3=F_3d_3=(0.8N)(0.6m-0.5m)$
$W_3=0.08J$
Now the total work done is given as
$W=0.06+0.1+0.08=0.24J$
