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3 years ago

Air bags greatly reduces the chance of injury in a car accident.explain how they do so in terms of energy transfer

Physics

1 answer:

VARVARA [1.3K]3 years ago

4 0

Answer:

Airbags reduce chances of injury by absorbing most of the impact force from the body during a car crash

Explanation:

In a car collision, the speed of the vehicle is suddenly bought to rest. All the kinetic energy is suddenly converted into other forms of energy.

The body of the driver keeps travelling forward under his inertia force due to his mass until he is slammed against the steering wheel. The steering wheel is a very rigid component, and so when the body slams against it, the body takes the deformation, absorbing some of the energy of the moving car. This sudden impact of energy can be fatal enough to gravely injure the driver because the body does not undergo much deformation. When an airbag is used, the crash automatically triggers the release of the airbag. Instead of the body colliding against the rigid steering wheel, it is now collided against the soft air bag. The airbag is very collapsible, and some of the kinetic energy of the car on the driver is converted into the deformation energy used to deform the airbag when they collide. In the process of deformation, the time of impact is extended, reducing the force impacted on the driver, reducing the fatality of the impact.

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$F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N$

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$F=k\Delta x$

where k is the spring constant and $\Delta x$ is the stretching of the spring. Since we know $\Delta x=2.92 cm=0.0292 m$ , we can re-arrange the equation to find the spring constant:

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In this second case, the force applied to the spring will be different, since the weight of the new object is different:

$F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N$

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

$\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm$

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance $\Delta x$ is equal to the elastic potential energy stored by the spring, given by:

$W=U=\frac{1}{2}k\Delta x^2$

Substituting k=907.5 N/m and $\Delta x=8.80 cm=0.088 m$ , we find the amount of work that must be done:

$W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J$

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Answer:

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