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A car's bumper is designed to withstand a 4.32 km/h (1.2 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 (in N) on a bumper that collapses 0.150 m while bringing a 820 kg car to rest from an initial speed of 1.2 m/s. (Enter a number.)

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Answer:

3936 N

Explanation:

m = Mass of car = 820 kg

u = Initial velocity = 1.2 m/s

v = Final velocity = 0

d = Distance = 0.15 m

The work done is given by the change in the kinetic energy of the system

[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow Fd=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow F=\dfrac{\dfrac{1}{2}m(v^2-u^2)}{d}\\\Rightarrow F=\dfrac{\dfrac{1}{2}820(0^2-1.2^2)}{0.15}\\\Rightarrow F=-3936\ N[/tex]

The magnitude of force is 3936 N

Answer:

[tex]F=3936\ N[/tex]

Explanation:

Given:

  • initial speed of the car, [tex]u=1.2\ m.s^{-1}[/tex]
  • mass of the car, [tex]m=820\ kg[/tex]
  • final speed of the car, [tex]v=0\ m.s^{-1}[/tex]
  • distance collapsed due to collision, [tex]s=0.15\ m[/tex]

Now by the law of energy conservation, the work done is the energy compensated in bringing the car to rest.

[tex]W=\Delta KE[/tex]

[tex]F.s=\frac{1}{2} m.u^2-\frac{1}{2} m.v^2[/tex]

where:

[tex]W=[/tex] work done by the force F

[tex]\Delta KE=[/tex] change in kinetic energy

[tex]F\times 0.15=\frac{1}{2}\times 820\times 1.2^2-0[/tex]

[tex]F=3936\ N[/tex]

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