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SVETLANKA909090 [29]

3 years ago

11

John is diving off a springboard which is 10 meters above the water. He weighs 75 kg. Based on this information, how fast will J

ohn be moving when he hits the water below?

1 answer:

Dennis_Churaev [7]3 years ago

6 0

Answer:

v = 14 m /s

Explanation:

vertical displacement h = 10 m

acceleration = g = 9.8 m /s²

initial vertical velocity u = 0

v² = u² + 2 g h

= 0 + 2 x 9.8 x 10

v = 14 m /s

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Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is

marshall27 [118]

Answer:

$x=0.0049\ m= 4.9\ mm$

$d=0.01153\ m=11.53\ mm$

Explanation:

Given:

mass of the object, $m=0.75\ kg$

elastic constant of the connected spring, $k=150\ N.m^{-1}$

coefficient of static friction between the object and the surface, $\mu_s=0.1$

(a)

Let x be the maximum distance of stretch without moving the mass.

<em>The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.</em>

$f=k.x$

$\mu_s.N=k.x$

where:

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(b)

Now, according to the question:

Amplitude of oscillation, $A= 0.0098\ m$

coefficient of kinetic friction between the object and the surface, $\mu_k=0.085$

Let d be the total distance the object travels before stopping.

<em>Now, the energy stored in the spring due to vibration of amplitude:</em>

$U=\frac{1}{2} k.A^2$

<u><em>This energy will be equal to the work done by the kinetic friction to stop it.</em></u>

$U=F_k.d$

$\frac{1}{2} k.A^2=\mu_k.N.d$

$0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d$

$d=0.01153\ m=11.53\ mm$

<em>is the total distance does it travel before stopping.</em>

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