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mihalych1998 [28]

3 years ago

9

Un auto de 1000 kg que lleva una velocidad de 72 km/h se detiene en una distancia de 100 metros cuando se le aplican los frenos.

(a) ¿Cuál es la fuerza constante producida por los fre. Nos? (b) Con esta fuerza, ¿cuál seria la distancia de frena do si la velocidad hubiera sido de 36 km/h?

1 answer:

Flura [38]3 years ago

6 0

Answer:

a) F = 2000 N , B) x = 25 m

Explanation:

a) To solve this exercise we can use the relationship between work and kinetic energy

W = ΔK

The job is

W = -F x

the negative sign is because the force of the brakes is contrary to the movement

as the car stops its final kinetic energy is zero

K = ½ m v²

let's substitute

- F x = 0 - ½ m v²

F = ½ m v² / x

Let's reduce the magnitudes to the SI system

v = 72 km / h (1000m / 1km) (1 h / 3600 s) = 20 m / s

let's calculate

F = ½ 1000 20²/100

F = 2000 N

b) x = ½ mv2 / F

let's slow down to the SI system

v = 36 km / h = 10 m / s

let's calculate

x = ½ 1000 10²/2000

x = 25 m

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<span>From the problem alone we can say that the dart and the block of wood combined into a single object moving together at the end. With that clue we know that the collision is an inelastic collision. The formula of an inelastic collision is:

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First let us sort out our given:Mass should be in kg to get the proper answer. Now let's assign m1 as the mass of the dart and m2 as the mass of the block.
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$22.6g$ x $\frac{1kg}{1000g}= 0.0226kg$

So now we settled that we can set our given as:
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dy = -1.5 m

a = 9.8m/s^2

viy = 0 (take note that the initial vertical velocity is 0)

t = ?

<span>We can use the UAM equations to solve for the time in the y-dimension (vertical) to get the horizontal velocity.

$dy = v_{iy}t + \frac{1}{2} at^{2}$ </span>

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<span> $1.5 = \frac{1}{2} (9.8)t^{2}$

$\frac{(2)(1.5)}{9.8}=t^{2}$

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$\sqrt{0.3061} = \sqrt{t^{2}$

$0.553s = t$

Now using this, we can get the horizontal (x-dimension) velocity using the formula:
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$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

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