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

Answer 1

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)

Answer 2

I don't understand it

Answer 3

Pls 🙏🙏 explain better

Answer 4

0.8 divided by 0.005 is 160

1.5+3.5N is 5

5x1N is 5

5 divided by 160 is 1.5219

Answer 5

Ajisefinni, I gtg, I was just letting you know, just in case you were going to ask anymore questions, bye.

Answer 6

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