Question

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behaves like a simple spring. (a) Calculate how much it would stretch if the same person were lying in it. (b) How much would it stretch if the person jumped from 38 m?

Answer 1

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

Answer 2

A) If the person is lying on the net, the net would stretch by : 0.0679 m

B) If the person jumped from 38m, the net would stretch by : 1.61 m

Given data :

Mass of person (m) = 62 kg

Initial height ( H₁ ) = 20 m

Net stretch ( H$_{f}$ ) = -1.2 m

Initial potential energy before jump ( PE₁ ) = mgH₁ = 62 * 9.81 * 20 = 12164.4J

Potential energy after the jump ( PE₂ ) = mgH$_{f}$ = 62 * 9.81 * ( -1.2 ) = -729.86J

The potential spring constant ( SE ) = 1/2 kx²

                                                           = 1/2 * k * ( 1.2 )²

Applying the principle of energy conservation

PE₁ = PE₂ + SE

12164.4 = - 729.86 + 0.5 * k * 1.44

∴ k = ( 12164.4 + 729.86 ) / ( 0.5 * 1.44 )

     = 17908.69

A) Calculate the amount of stretch if same person lies on the net

1/2 kx² = mgx  ---- ( 1 )

Where x = stretch

equation ( 1 ) becomes equation ( 2 )

1/2 kx = mg ----- ( 2 )

x ( amount of stretch ) = ( 2mg ) / k

                                     = ( 2 * 62 * 9.81 ) / 17908.69

                                     = 0.0679 m

B ) Calculate How much the net would stretch if Height = 38 m

Hi = 38 m

MgHi = mgH$_{f}$ + 1/2 kx² ----- ( 3 )

where ; H$_{f}$ = -x

Back to equation ( 3 )

62 * 9.81 * 38 = 62 * 9.81 * (-x) + 1/2 ( 17908.69 x² )

23112.36 + 608.22x - $\frac{17908.69 x^{2} }{2}$ = 0

8954.345 x² - 608.22x - 23112.36 = 0 ----- ( 4 )

Resolving the quadratic equation ( 4 )

x = ± 1.60623 ≈ 1.61 m

Therefore we can conclude that the net would stretch 1.61 m if the person jumped from 38 m

Hence we can conclude that If the person is lying on the net, the net would stretch by : 0.0679 m, and If the person jumped from 38m the net would stretch by : 1.61 m

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