Asked by Ajisefinni Oluwatosin

A spring of natural length 1.5m is extended 0.005m by a force of 0.8Newton what will it length be when the applied force is 3.2Newton
Answered by Shark_boii
k = 0.8N/0.0050m = 160 N/m. L = 1.5 + 3.5N*1m/160N. = 1.5219 m.

(Hope this helps and also let me know if you can't understand this)
Answered by Ajisefinni Oluwatosin
I don't understand it
Answered by Ajisefinni Oluwatosin
Pls 🙏🙏 explain better
Answered by Shark_boii
0.8 divided by 0.005 is 160

1.5+3.5N is 5

5x1N is 5

5 divided by 160 is 1.5219
Answered by Shark_boii
Ajisefinni, I gtg, I was just letting you know, just in case you were going to ask anymore questions, bye.
Answered by Anonymous
3.2 is four times 0.8
so it will be extended four times as far
0.020
plus original length 1.5
= 1.520

3.2 Newtons Shark, not 3.5
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