Respuesta :
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]
The net force acting on the car is equal to 700 Newton.
Given the following data:
- Mass of car = 1400 kg
- Initial speed = 18 m/s
- Final speed = 14 m/s
- Displacement = 35.0 m
To calculate the net force acting on the car, we would apply the work-energy theorem:
The work-energy theorem
According to the work-energy theorem, the amount of work done by the car must balance the kinetic energy possessed by the car due to its motion.
From the work-energy theorem, we have:
[tex]Work\;done = \Delta K.E\\\\Work\;done = F_{net}\times d=\frac{1}{2} M(V^2-U^2)[/tex]
The change in kinetic energy.
[tex]\Delta K.E = \frac{1}{2} M(V^2-U^2)\\\\\Delta K.E = \frac{1}{2} \times 1400 \times (18^2-14^2)\\\\\Delta K.E = 700 \times (324-289)\\\\\Delta K.E = 700 \times 35\\\\\Delta K.E = 24500\;Joules[/tex]
Now, we can calculate the net force:
[tex]F_{net} = \frac{\Delta K.E}{displacement} \\\\F_{net} = \frac{24500}{35}[/tex]
Net force = 700 Newton.
Read more on work-energy theorem here: https://brainly.com/question/22236101
