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3 years ago

Man made objects that spread an impulse over a large amount of time

Physics

1 answer:

lyudmila [28]3 years ago

5 0

Answer:

Examples of man-made objects that spread an impulse over a large amount of time are "airbags" in vehicles and "arrestor beds" (for trucks).

Explanation:

The question above is highly related to the topic about "Impulse" in Physics.

"Impulse" refers to an object's change in momentum (the amount of motion in an object) when a force acts upon it for an interval time. When it comes to providing safety to people when it comes to vehicular crashes, impulse plays a vital role.

Let's take the example of airbags in vehicles. Once a vehicle collides with another object, the driver is carried by a forward motion. Without airbags, the time is normally shorter for the driver to be stopped by the windshield. This results to a greater force. However, with the presence of air-bags, the driver will hit the airbag, instead of the windshield. This will lengthen the time of the impact, thus reducing the force.

Another example are the arrestor beds for trucks. Arrestor beds have been designed in order for trucks to stop, since it's hard to maneuver them. With the help of arrestor beds, trucks are able to come to a stop with a longer time interval, but decreased force.

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A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori

olasank [31]

Explanation:

The given data is as follows.

k = 130 N/m, $\Delta x$ = 17 cm = 0.17 m (as 1 m = 100 cm)

mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

Energy = $\frac{1}{2}kx^{2}$

Now, spring energy gets converted into kinetic energy when the box is launched.

So, $\frac{1}{2}kx^{2}$ = $\frac{1}{2}mv^{2}$

$\frac{1}{2} \times 130 \times (0.17)^{2}$ = $\frac{1}{2} \times 2.8 \times v^{2}$

$v^{2} = \frac{3.757}{2.8}$

= 1.34

v = 1.15 m/sec

Now,

Frictional force = $\mu \times mg$

= $0.15 \times 2.8 \times 9.8$

= 4.116 N

Also, Kinetic energy = work done by friction

$\frac{1}{2}mv^{2} = F_{f} \times d$

$\frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d$

1.8515 = $4.116 \times d$

d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

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