Answer:
The change in kinetic energy of the car is -189434 joules.
The work done by the car is 189434 joules.
The magnitude of the force that pushed the front of the car is 498510.526 newtons.
Explanation:
Given that the car is moving on a horizontal ground, by the Principle of Energy Conservation and the Work-Energy Theorem we get the following identity:
[tex]\Delta K+\Delta W = 0[/tex] (1)
Where:
[tex]\Delta K[/tex] - Change in the translational kinetic energy of the car, measured in joules.
[tex]\Delta W[/tex] - Work done by the car, measured in joules.
By applying the defintions of translational kinetic energy and work, we expand and simplify the resulting equation:
[tex]\frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})+F\cdot \Delta s = 0[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the car, measured in meters per second.
[tex]F[/tex] - Force exerted by the car, measured in newtons.
[tex]\Delta s[/tex] - Travelled distance of the front of the car, measured in meters.
The change in the kinetic energy of the car and the work done by the car are, respectively: [tex](v_{o} = 14\,\frac{m}{s}, v_{f}= 0\,\frac{m}{s}, m = 1933\,kg, \Delta s = 0.38\,m)[/tex]
Translational kinetic energy
[tex]\Delta K = \frac{1}{2}\cdot m\cdot (v_{f}^{2}-v_{o}^{2})[/tex]
[tex]\Delta K = \frac{1}{2}\cdot (1933\,kg)\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(14\,\frac{m}{s} \right)^{2}\right][/tex]
[tex]\Delta K = -189434\,J[/tex]
Work done by the car
[tex]\Delta W = -\Delta K[/tex]
[tex]\Delta W = 189434\,J[/tex]
Magnitude of the force
[tex]F = \frac{\Delta W}{\Delta s}[/tex]
[tex]F = \frac{189434\,J}{0.38\,m}[/tex]
[tex]F = 498510.526\,N[/tex]
The change in kinetic energy of the car is -189434 joules.
The work done by the car is 189434 joules.
The magnitude of the force that pushed the front of the car is 498510.526 newtons.
