Answer
The required net force to stop the car on a dime is $3.69\times 10^6~N$.
Work Step by Step
We can convert the speed to units of m/s.
$v = (45.0~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 12.5~m/s$
We can find the rate of deceleration of the car.
$a = \frac{v^2-v_0^2}{2x} = \frac{0-(12.5~m/s)^2}{(2)(0.018~m)}$
$a = -4340~m/s^2$
The magnitude of the acceleration is $4340~m/s^2$. We can use a force equation to find the net force.
$\sum F = ma$
$\sum F = (850~kg)(4340~m/s^2)$
$\sum F = 3.69\times 10^6~N$
The required net force to stop the car on a dime is $3.69\times 10^6~N$.
