Answer
(a) $-0.11MJ$
(b) heat and sound
Work Step by Step
(a) We know that
$\Delta K.E=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$
$\implies \Delta K.E=\frac{1}{2}m(v_f^2-v_i^2)$
We plug in the known values to obtain:
$\Delta K.E=\frac{1}{2}(1300)((11)^2-(17)^2)$
$\Delta K.E=-1.1\times 10^5J=-0.11MJ$
(b) The missing kinetic energy is actually converted into heat and sound when the brakes are applied and friction is produced.
