if a force of 100N stretches a spring by 0.1cm find.

I need the answer now please

Look at David's question below this one.

then work done = (1/2) k x^2

use that k
x = 0.003 meters for answer in Joules

a. k = 100N/0.1cm = 1000N/cm.

*
b. F = 1000N/cm * 0.3cm = 300 N.
W = F*d = 300*0.003 = 0.9 Joules.

Henry, no, F not constant

F=ke

100=k*(0.1/100)
0.1k =10000
K=100000N/m

W=1/2ke²
W=1/2*100000*0.3²
W=(100000*0.09)/2
W=9000/2
W=4500J

F = 100n

E= 0.1cm= 0.001m
F = ke = 100÷0.001= 100000
E= 0.3cm = 0.003m
W= 1/2 fe = 1/2 ke ^2
W= 1/2 × 100000x 0.003^2
W= 0.45J

thank you the answer was perfect

4.0

What do you mean by "4.0"?

From hooke's law

F=ke therefore k=fe=100/0.001=100000N/m-1
The workdone is w=1/2ke^2
=1/2×100000×(0.03)^2=0.45j

Thank you for the clarification. The elastic constant (k) is 100000 N/m and the work done in stretching the spring by 0.3cm is 0.45 J.