Question

(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).

Short answer

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

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

Step-by-step solution

Definition of Concept

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,

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

Find the force needed to bring the car into motion

(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 \(2474{\rm{ N}}\).

Comparing forces

(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.