Physics, 04.02.2020 04:58 Raekwon3232

You have a light spring which obeys hooke's law. this spring stretches 2.92 cm vertically when a 2.70 kg object is suspended from it. determine the following. (a) the force constant of the spring (in n/m), (b) the distance (in cm) the spring stretches if you replace the 2.70 kg object with a 1.35 kg object, (c) the amount of work (in j) an external agent must do to stretch the spring 8.80 cm from its unstretched position

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Answer from: hailekelly1864

The force constant of the spring is 906.2 N/m.

The extension of the spring when 2.7 kg is replaced with 1.35 kg is 2.896 cm.

The work an external agent must do to stretch the spring by 8.8 cm is 3.51 J.

The given parameters;

extension of the spring, x = 2.92 cm = 0.0292 mmass of the object, m = 2.7 kg

The weight of the object suspended on the spring;

W = mg

W = 2.7 x 9.8 = 26.24 N

The force constant of the spring is calculated as;

$F= kx\\\\k = \frac{F}{x} \\\\k = \frac{26.46}{0.0292} \\\\k = 906.2 \ N/m$

The extension of the spring when 2.7 kg is replaced with 1.35 kg;

$F = kx\\\\x = \frac{F}{k} \\\\x = \frac{26.24}{906.2} \\\\x = 0.0289 \ m\\\\x = 2.896 \ cm$

The work an external agent must do to stretch the spring by 8.8 cm

$W = \frac{1}{2} kx^2\\\\W = \frac{1}{2} \times 906.2 \times (0.088)^2\\\\W = 3.51 \ J$

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Answer from: erik118

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N

The spring obeys Hook's law:

$F=k\Delta x$

where k is the spring constant and $\Delta x$ is the stretching of the spring. Since we know $\Delta x=2.92 cm=0.0292 m$ , we can re-arrange the equation to find the spring constant:

$k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m$

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

$\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm$

The amount of work that must be done to stretch the string by a distance $\Delta x$ is equal to the elastic potential energy stored by the spring, given by:

$W=U=\frac{1}{2}k\Delta x^2$

Substituting k=907.5 N/m and $\Delta x=8.80 cm=0.088 m$ , we find the amount of work that must be done:

$W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J$

Answer from: Quest

Depotion can i think

Answer from: Quest

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

so, refractive index of medium = 300000/200000 = 1.5

1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

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You have a light spring which obeys hooke's law. this spring stretches 2.92 cm vertically when a 2.7...