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garik1379 [7]
2 years ago
5

11 An unstretched spring is 12,0 cm long. A load of 5.0N stretches it to 15.0cm. How long will it be under a load

Physics
1 answer:
Rom4ik [11]2 years ago
4 0

Answer: Approximately 22 cm

=========================================================

Explanation:

The unstretched spring is 12.0 cm long. When adding a load of 5.0 N, it stretches to 15.0 cm. This is a displacement of 15.0 - 12.0 = 3.0 cm, which is the amount the spring is stretched.

Convert this displacement to meters (so that it fits with the meters unit buried in Newtons).

3.0 cm = (3.0)/100 = 0.03 m

Apply Hooke's Law to find the spring constant k

F = -kx

5.0 = -k*(0.03)

k = -(5.0)/(0.03)

k = -166.667 approximately

Now we must find the displacement x when F = 15 newtons

F = -kx

-kx = F

x = F/(-k)

x = -F/k

x = -15/(-166.667)

x = 0.089 approximately

x = 0.1

The displacement to one decimal place is about 0.1 meters, which converts to 100*0.1 = 10 cm

So the spring will be stretched to about 12cm+10cm = 22 cm

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y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

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y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

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v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

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v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

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2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

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