F=kx (Hooke's Law) Problem

Can someone guide me through this problem:

2. A mass of 500 g stretches a spring 8.0 cm when it is attached to it.

What additional weight would you have to add to it so that the spring is stretched 10 cm?

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I tried plugging in numbers into the rule and that didn't lead me anywhere!

Easiest way using canadian teaching method

First easiest way to do it is to first get all your units in their easiest form in canada on the end of year exam they usually ask you use kg-m-N

so

500g becomes .5kg -- F=mg F=.5(9.8) = 4.9N

80cm becomes .08m

so now using the rule F=kx you need

Force in N=force constant N/m * displacement in m

4.9N = k * .08m

k=4.9/.o8

k=61.25

It is the same spring so same K value, we are now given a new displlacement

k=61.25 N/m

x=10cm = .1m

With this info we can find the Force

F=kx

F=(61.25)(10)

F=6.125N

Now that we have the new force you can get the total weight

F=mg

6.125=m(9.8)

m=.625

that is the total weight they asked you how much was added on so just take your total weight and substract your initial weight

.625-.5= .125kg which is your answer

and thats that. the way your expected to do it on governement exam in june (Wink)