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A car's bumper is designed to withstand a 6.48-km/h (1.8-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.210 m while bringing a 830-kg car to rest from an initial speed of 1.8 m/s.

A cars bumper is designed to withstand a 648kmh 18ms collision with an immovable object without damage to the body of the car The bumper cushions the shock by a class=

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Answer:

6402.86 N

Explanation:

taking into account the energy-work theorem, we can write the following equation

[tex]\begin{gathered} W=K_f-K_i \\ Fd=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \end{gathered}[/tex]

Where F is the force, d is the distance, m is the mass, vf is the final velocity and vi is the initial velocity.

Solving for F, we get:

[tex]F=\frac{mv_f^2-mv_i^2}{2d}[/tex]

Replacing m = 830 kg, vf = 0 m/s, vi = 1.8 m/s, and d = 0.210 m, we get:

[tex]F=\frac{(830)(0)^2-(830)(1.8)^2}{2(0.210)}=-6402.86\text{ N}[/tex]

Therefore, the magnitude of the force is 6402.86 N