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Fantom [35]
3 years ago
5

A car's bumper is designed to withstand a 6.84 km/h (1.9-m/s) collision with an immovable object without damage to the body of t

he 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 0.285 m while bringing a 830 kg car to rest from an initial speed of 1.9 m/s.
Physics
1 answer:
den301095 [7]3 years ago
3 0

Answer:

F = 5.256 x 10^{3} 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:

W = F.d, ΔK =\frac{1}{2} mv^{2}_{f}   - \frac{1}{2} mv^{2}_{i}  , F.d = \frac{1}{2}mv^{2}_{f} -\frac{1}{2} mv^{2}_{i}

where:

W = work done by the force

F = Force

d = Distance travelled

m = Mass of the car

vf, vi = final and initial velocity of the car

kf, ki = final and initial kinetic energy of the car

Given the parameters;

m = 830kg

vi = 1.9 m/s

vf = 0 km/h

d = 0.285 m

Inserting the information we have:

F.d = \frac{1}{2} mv^{2}_{f}   - \frac{1}{2} mv^{2}_{i}

F = \frac{\frac{1}{2} mv^{2}_{f}   - \frac{1}{2} mv^{2}_{i} }{d}

F = \frac{ 0   - \frac{1}{2}  X830 X 1.9^{2} }{0.285}

F = 5.256 x 10^{3} N

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Answer:

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Explanation:

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Why?

We can find the distance between each pair of points using the following formula:

d(P_1,P_2)=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

So, calculating we have:

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d(E,F)=\sqrt{(-2-(-2))^{2}+(-5-(-1))^{2}}\\\\d(E,F)=\sqrt{0+(-5+1)^{2}}=\sqrt{0+(-4)^{2}}=\sqrt{16}=4units

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d(A,B)=\sqrt{(5-(5))^{2}+(-2-(4))^{2}}\\\\d(A,B)=\sqrt{0+(-6)^{2}}=\sqrt{0+36}=\sqrt{36}=6units

Have a nice day!

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