Answer
(a) $85~N$
(b) Increased.
Work Step by Step
(a) $1~mi=1609~m$, $1~h=3600~s$
$\Delta x=1.7~m$, $v_0=0$, $m=0.15~kg$
$v=98~mi/h=(98~mi/h)(\frac{1609~m}{1~mi})(\frac{1h}{3600~s})=43.8~m/s$
$v^2=v_0^2+2a\Delta x$
$(43.8~m/s)^2=0^2+2a(1.7~m)$
$a=\frac{(43.8~m/s)^2}{3.4~m}=564~m/s^2$
$F=ma=(0.15~kg)(564~m/s^2)=85~N$
(b) From the equation above, if the mass is increased the force required is increased as well.
