Answer
(a) The work done on the plane by the cables is $-5.04\times 10^7~J$
(b) The force exerted on the plane by the cables is $6.0\times 10^5~N$
Work Step by Step
(a) We can find the mass of the plane:
$mg = 220~kN$
$m = \frac{220~kN}{9.80~m/s^2}$
$m = 22,450~kg$
The work done by the cables on the plane is equal to the plane's change in kinetic energy. We can find the work done by the cables:
$Work = KE_2-KE_1$
$Work = 0-\frac{1}{2}mv^2$
$Work = -\frac{1}{2}(22,450~kg)(67~m/s)^2$
$Work = -5.04\times 10^7~J$
The work done on the plane by the cables is $-5.04\times 10^7~J$
(b) We can use the magnitude of the work done by the cables to find the magnitude of the force exerted on the plane by the cables:
$Work = F~d$
$F = \frac{Work}{d}$
$F = \frac{5.04\times 10^7~J}{84~m}$
$F = 6.0\times 10^5~N$
The force exerted on the plane by the cables is $6.0\times 10^5~N$
