Answer
(a) $0.76m/s$
(b) $0.148m$
Work Step by Step
(a) We can find the required speed as follows:
$W=F\Delta x$
We plug in the known values to obtain:
$W=(0.8N)(0.5m-0.27m)+(0.4N)(0.75m-0.5m)+(0.2N)(0.99m-0.5m)$
$W=0.332J$
$\implies W=F\Delta x=0.332J$
We know that
$W=\Delta K.E$
$W=\frac{1}{2}m(v_f^2-v_i^2)$
$\implies F\Delta x=\frac{1}{2}m(v_f^2-v_i^2)$
This simplifies to:
$v_f=\sqrt{\frac{2F\Delta x}{m}+v_i^2}$
We plug in the known values to obtain:
$\implies v_f=\sqrt{\frac{2(0.332J)}{1.7Kg}+(0.44m/s)^2}$
$v_f=0.76m/s$
(b) We can find the required location as follows:
$F\Delta x=(0.8N)(-0.02m)=-0.016J$ and $x=0.25m$
and $v_f=\sqrt{\frac{2(-0.016J)}{1.7Kg}+(0.44m/s)^2}$
$v_f=0.41806m/s$
We know that
$\frac{1}{2}m(v_f^2-v_i^2)=F(x_f-x_i)$
This simplifies to:
$x_f=x_i+\frac{m}{2F}(v_f^2-v_i^2)$
We plug in the known values to obtain:
$x_f=(0.25m)+\frac{1.7Kg}{(2)(0.6N)}[(0.32m/s)^2-(0.41806m/s)^2]$
$x_f=0.148m$
