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marusya05 [52]
3 years ago
5

If the same fish is attached to the end of the unstretched spring and then allowed to fall from rest, through what maximum dista

nce does it stretch the spring? (Hint: Calculate the force constant of the spring in terms of the distance d and the mass m of the fish.)
Physics
1 answer:
andrew11 [14]3 years ago
3 0

Answer:

x=2d

Explanation:

initial stretch in the spring is d

so using Hook's law

at equilibrium position

k×d=mg

where k= spring constant

m= mass of fish

g= acceleration due to gravity.

d=mg/k ................ (1)

in second case  by energy conservation

1/2 kx^2=mgx

x=2mg/k

using equation 1

x=2d

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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 behav
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Answer:

a) x = 0.098

b) x = 2.72 m

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(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:

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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

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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

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