Question

1. a tractor - trailer is traveling at a speed of \\(\frac{15.64mathrm{m}}{mathrm{s}}\\) and needs to come to a complete stop at a stoplight. the mass of the truck is 36,287.4 kg. how much work is done by the tractor - trailers brakes to stop it?\\(\circ - 4,438,123.2mathrm{j}\\)\\(\circ 4.4mathrm{j}\\)\\(\circ - 244.6mathrm{j}\\)\\(\circ 8,876,246.4mathrm{j}\\)

Explanation:

Step1: Recall work-kinetic energy theorem

Work done equals change in kinetic energy: $W = \Delta KE = KE_f - KE_i$

Step2: Define final and initial KE

Final velocity $v_f=0$, so $KE_f=0$. Initial $KE_i=\frac{1}{2}mv_i^2$

Step3: Substitute values into formula

$m=36287.4\ \text{kg}$, $v_i=15.64\ \text{m/s}$
$KE_i=\frac{1}{2} \times 36287.4 \times (15.64)^2$

Step4: Calculate initial KE

First compute $(15.64)^2 = 244.6096$
Then $\frac{1}{2} \times 36287.4 \times 244.6096 = 4438123.2$

Step5: Solve for work done

$W = 0 - 4438123.2 = -4438123.2\ \text{J}$

Answer:

-4,438,123.2 J