Answer:
8512 N
Explanation:
From the work energy theorem we know that: The net work done on a particle equals the change in the particles kinetic energy:
[tex]W=\Delta K=K_{f}-K_{i} \\ \\qquad \begin{array}{r} W=F \cdot d, \Delta K=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\ F \cdot d=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}[/tex]
Where:
-W is the work done by the force.
- F is the force actin on the.
- d is the distance travelled.
- m is the mass of the car.
- [tex]v_{f}, v_{i}[/tex] are the final and the initial velocity of the car
[tex]K_{f}, K_{i}[/tex] are the final and the kinetic energy of the car.
Givens: [tex]m=830 \mathrm{~kg}, v_{i}=2 \mathrm{~m} / \mathrm{s}, v_{f}=0 \mathrm{~km} / \mathrm{h}, d=0.195 \mathrm{~m}[/tex]
Plugging known information to get:
[tex]F \cdot d &=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\ F &=\frac{\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}}{d} \\ &=\frac{0-\frac{1}{2} \times 830 \times 2^{2}}{0.195} \\ &=8.512 \times 10^{3} \\ F &=8.512 \times 10^{3} \mathrm{~N}[/tex]