Answer:
Net work done, [tex]W=3.08\times 10^5\ J[/tex]
Explanation:
It is given that,
Mass of the plane, [tex]m=7.5\times 10^3\ kg[/tex]
Acceleration of the plane, [tex]a=1.2\ m/s^2[/tex] (upwards)
Distance covered, d = 34.3 m
We need to find the net work done on the plane as it accelerates upward. The product of force and the distance covered is equal to work done. It is given by :
[tex]W=F\times d[/tex]
[tex]W=m\times a\times d[/tex]
[tex]W=7.5\times 10^3\ kg \times 1.2\ m/s^2 \times 34.3\ m[/tex]
W = 308700 J
or
[tex]W=3.08\times 10^5\ J[/tex]
So, the net work done on the plane is [tex]3.08\times 10^5\ J[/tex]. Hence, this is the required solution.