the spring stretched by x=5 cm
F = kx
so now you can find k
work = 1/2 kx^2
watch the units
A spring of natural length 30cm is stretched to 35cm by a load of 15N. Calculate the extension produced,the elastic constant of the spring,the work done in stretching the spring.
3 answers
a. d = 35-30 = 5 cm stretched.
b. k = F/d = 15/5 = 3N/cm = 300N./m.
c. W = F*d = 15*0.05 = 0.75 J.
b. k = F/d = 15/5 = 3N/cm = 300N./m.
c. W = F*d = 15*0.05 = 0.75 J.
No
1 answer