Answer:

The net force on the car is 2560 N.

Explanation:

According to work energy theorem, the amount of work done is equal to the change of kinetic energy by an object. If ‘[tex]W[/tex]’ be the work done on an object to change its kinetic energy from an initial value ‘[tex]K_{i}[/tex]’ to the final value ‘[tex]K_{f}[/tex]’, then mathematically,

[tex]W = K_{f} – K_{i} = \dfrac{1}{2}~m~(v_{f}^{2} – v_{i}^{2})………………………………….(I)[/tex]

where ‘[tex]m[/tex]’ is the mass of the object and ‘[tex]v_{i}[/tex]’ and ‘[tex]v_{f}[/tex]’ be the initial and final velocity of the object respectively. If ‘[tex]F_{net}[/tex]’ be the net force applied on the car, as per given problem, and ‘[tex]s[/tex]’ is the displacement occurs then we can write,

[tex]W = F_{net}~.~s……………………………………………….(II)[/tex]

Given, [tex]m = 1400~Kg,~v_{i} = 18~m~s^{-1}~v_{f} = 14~m~s^{-1}~and~s = 35~m[/tex].

Equating equations (I) and (II),

[tex]&& – F_{net} \times 35~m = \dfrac{1}{2} \times 1400~Kg~\times(14^{2} – 18^{2})~m^{2}~s^{-2}\\&or,& F_{net} = \dfrac{\dfrac{1}{2} \times 1400~Kg~\times(14^{2} – 18^{2})}{35}~N\\&or,& F_{net} = 2560~N[/tex]