Answer
(a) $-29.4MJ$
(b) $5.51MJ$
(c) not conserved
Work Step by Step
(a) We know that
$\Delta U=mgh_2-mgh_1$
We plug in the known values to obtain:
$\Delta U=(15800Kg)(9.8m/s^2)(1440m-1630m)$
$\implies \Delta U=-29.4MJ$
(b) We can find the required kinetic energy as follows:
$\Delta K.E=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$
We plug in the known values to obtain:
$\Delta K.E=\frac{1}{2}(1580Kg)[(29m/s)^2-(12m/s)^2]$
$\Delta=5.51MJ$
(c) We know that the total kinetic energy of the system is conserved when
$E_i=E_f$
Now $E_i=U+K.E$
$\implies E_i=mgh_1+\frac{1}{2}mv_i^2$
We plug in the known values to obtain:
$E_i=(15800Kg)[(9.8m/s^2)\times (1630m)+\frac{1}{2}(12m/s)^2]=2.53\times 10^8J$
and $E_f=mgh_2+\frac{1}{2}mv_f^2$
We plug in the known values to obtain:
$E_f=(15800Kg)[(9.8m/s^2)\times(1440m)+\frac{1}{2}(29m/s)^2]$
$E_f=2.3\times 10^8J$
We can see that $E_i\neq E_f$; hence, the mechanical energy of the truck is not conserved.
