The driver of a 1024 kg truck traveling 32.6 m/s North on I-77 has to make a sudden stop. If it...

The driver of a 1024 kg truck traveling 32.6 m/s North on I-77 has to make a sudden stop. If it requires a distance of 57.1 m for the truck to stop, use conservation of energy to find the net force acting on the truck.

Mass of the truck: {eq}\color{blue}{m = 1024\, \rm kg} {/eq}.

Initial velocity of the truck: {eq}\displaystyle{\color{blue}{v_0 = 32.6\, \rm m/s}} {/eq}

Final velocity of the truck: {eq}\color{blue}{v = 0\, \rm m/s} {/eq}

Therefore, total change in kinetic energy is given by:

{eq}\displaystyle{\begin{align*} \color{blue}{\Delta KE} &= \color{blue}{\frac{1}{2}mv^{2} - \frac{1}{2}mv^{2}_0}\\ &= \color{blue}{-\frac{1}{2}mv^{2}_0\, \, \, \, \, \, (\because v = 0)}\\ \end{align*}} {/eq}.

Let net force acting on the truck in the direction of its displacement be {eq}\color{blue}{F} {/eq}.

Displacement of the truck: {eq}\color{blue}{s = 57.1\, \rm m} {/eq}.

Therefore, work done by the force is given by:

{eq}\displaystyle{\color{blue}{W = F\, s}} {/eq}

Therefore, by Work-Energy Theorem, we write:

{eq}\displaystyle{\color{blue}{W = \Delta KE}\\ \implies \color{blue}{F\, s = -\frac{1}{2}mv^{2}_0}} {/eq}

Therefore, net force acting on the truck can be given as:

{eq}\displaystyle{\begin{align*} \color{red}{F} &= \color{blue}{-\frac{mv^{2}_0}{2\, s}}\\ &= \color{blue}{-\frac{1024\times (32.6)^{2}}{2\times 57.1}}\\ &\approx \color{red}{-9,529.48\, \rm N}\\ \end{align*}} {/eq}

Negative sign indicates that net force acting on the truck as a retarding force is opposite in direction to displacement of the truck.