(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop).

(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a).

Work energy theorem: According to the work-energy theorem, the work done by the body equals the change in kinetic energy of the body.

Mathematically,

\(W = K{E_f} - K{E_i}\)

(a)

From work-energy theorem,

\(\begin{aligned}W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2}\\Fd = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2}\end{aligned}\)

Here,\(F\)is the force required to bring the car into motion,\(m\)is the mass of car\(\left( {m = 950{\rm{ kg}}} \right)\),\(v\)is the final velocity of the car\(\left( {v = 90{\rm{ km}}/{\rm{h}}} \right)\),\(u\)is the initial velocity of the car (\(u = 0\)as the car was initially at rest),\(a\)is the acceleration of the car and\(d\)is the distance traveled by car\(\left( {d = 120{\rm{ m}}} \right)\).

The expression for the force is given as,

\(F = \frac{{m\left( {{v^2} - {u^2}} \right)}}{{2d}}\)

Putting all known values,

\(\begin{aligned}F &= \frac{{\left( {950{\rm{ kg}}} \right) \times \left( {{{\left( {90{\rm{ km}}/{\rm{h}}} \right)}^2} - {{\left( 0 \right)}^2}} \right)}}{{2 \times \left( {120{\rm{ m}}} \right)}}\\ &= \frac{{\left( {950{\rm{ kg}}} \right) \times \left( {{{\left\{ {\left( {90{\rm{ km}}/{\rm{h}}} \right) \times \left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right) \times \left( {\frac{{1{\rm{ hr}}}}{{3600{\rm{ s}}}}} \right)} \right\}}^2} - {{\left( 0 \right)}^2}} \right)}}{{2 \times \left( {120{\rm{ m}}} \right)}}\\ \approx 2474{\rm{ N}}\end{aligned}\)

Therefore, the force needed to bring car from rest to motion is \(2474{\rm{ N}}\).

(b)

From work-energy theorem,

\(\begin{aligned}W' = \frac{1}{2}m{{v'}^2} - \frac{1}{2}m{{u'}^2}\\F'd' = \frac{1}{2}m{{v'}^2} - \frac{1}{2}m{{u'}^2}\end{aligned}\)

Here,\(F'\)is the force exerted on car during collision,\(m\)is the mass of car\(\left( {m = 950{\rm{ kg}}} \right)\),\(v'\)is the final velocity of the car (\(v' = 0\)as the car stops),\(u'\)is the initial velocity of the car (\(u' = 90{\rm{ km}}/{\rm{h}}\)as the car was moving at its top speed), and\(d'\)is the distance travelled by the car while stopping\(\left( {d' = 2{\rm{ m}}} \right)\).

The expression for the force exerted on the car in course to collision is,

\(F' = \frac{{m\left( {{{v'}^2} - {{u'}^2}} \right)}}{{2d'}}\)

Putting all known values,

\(\begin{aligned}F' &= \frac{{\left( {950{\rm{ kg}}} \right) \times \left( {{{\left( 0 \right)}^2} - {{\left( {90{\rm{ km}}/{\rm{h}}} \right)}^2}} \right)}}{{2 \times \left( {{\rm{2 m}}} \right)}}\\& = \frac{{\left( {950{\rm{ kg}}} \right) \times \left( {{{\left( 0 \right)}^2} - {{\left\{ {\left( {90{\rm{ km}}/{\rm{h}}} \right) \times \left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right) \times \left( {\frac{{1{\rm{ hr}}}}{{3600{\rm{ s}}}}} \right)} \right\}}^2}} \right)}}{{2 \times \left( {{\rm{2 m}}} \right)}}\\ &= - 148437.5{\rm{ N}}\end{aligned}\)

As a result, the magnitude of the force required to stop the car is\(148437.5{\rm{ N}}\).

The ratio of force required to stop the car to the force required to set the car in motion is,

\(\begin{aligned}\frac{{F'}}{F} &= \frac{{148437.5{\rm{ N}}}}{{2474{\rm{ N}}}}\\\frac{{F'}}{F} &= 60\\F'& = 60F\end{aligned}\)

Therefore, the force exerted on the car to stop is \(60\) times the force needed to bring it in motion.