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A 1.0 kg weight suspended from a spring is pulled to 0.25 m below its equilibrium point.

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A 1.0 kg weight suspended from a spring is pulled to 0.25 m below its equilibrium point.

If the spring has a spring constant (k) of 50.0 N/m , at what rate will the mass accelerate when it is released?

200 m/s2

12.5 m/s2

6.5 m/s2

2.7 m/s2

i know forsure it isnt 2.7 because my teacher said it was incorrect?

Dec 3, 2019

#1
+21979
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Spring force = k l    k = 50 N/m

stretch it .25 m     .25 m x 50 N/ m = 12.5 Newtons

Now remember   f= ma ?     You have found f = 12.5   and you are given mass = 1 kg    ....you should be able to find 'a'

Dec 3, 2019
edited by ElectricPavlov  Dec 3, 2019

#1
+21979
+1

Spring force = k l    k = 50 N/m

stretch it .25 m     .25 m x 50 N/ m = 12.5 Newtons

Now remember   f= ma ?     You have found f = 12.5   and you are given mass = 1 kg    ....you should be able to find 'a'

ElectricPavlov Dec 3, 2019
edited by ElectricPavlov  Dec 3, 2019
#2
+1686
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so your saying the answer is 12.5 ? because 1*a = 12.5

jjennylove  Dec 3, 2019
#3
+109395
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F  = ma

12.5 Newtons  =  12.5kg * m  / s^2

12.5 kg m / s^2  =  1.0 kg  *  a        [ divide out kg ]

12.5 m /s^2   =  a

CPhill  Dec 3, 2019
#4
+1686
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for some reason, tihs had thrown me off :/

jjennylove  Dec 3, 2019
#5
+21979
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Now I am not so sure, JL....the spring mass was started at equalibrium......so the mg of gravity was already being balanced by the spring.....THEN you ADDED  .25 x 50   N  of upward force which is what the weight experiences when you release it...I think this is correct.... the a will be 12.5 m/s^2 .

The NET force will be upward due to the spring being further stretched......the spring at equalibrium has already balanced (i.e. cancelled out) the mg .....

Did your teacher tell you the correct answer was 2.7 m/s^2 ?

ElectricPavlov  Dec 3, 2019