extension is proportional to the force
5 N extends the spring 5 cm ... 1 N/cm
relaxed length is 20.0 cm
spring is 23.0 cm with a 3 N weight
The spiral spring of a spring balance is 25.0cm long when 5N hangs on it as and 30.0cm long when the weight is 10N
What is the length of the spring
If the weight is 3N assuming hooke's law is obeyed?
4 answers
k = (10N-5N)/(30cm-25cm) = 5N/5cm = 1N/cm . = spring constant.
Lo + (1cm/N * 5N) = 25.
Lo + 5 = 25,
Lo = 20 cm = Initial length.
L = Lo + (1cm/N * 3N) = 20 + 3 = 23 N. = Length with 3N-Wt.
Lo + (1cm/N * 5N) = 25.
Lo + 5 = 25,
Lo = 20 cm = Initial length.
L = Lo + (1cm/N * 3N) = 20 + 3 = 23 N. = Length with 3N-Wt.
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1 answer