Question
if a force of 100N stretches a spring by 0.1cm find.
a.the elastic constant
b.the work done in stretching the spring 0.3cm if the elastic limit is not exceeded
a.the elastic constant
b.the work done in stretching the spring 0.3cm if the elastic limit is not exceeded
Answers
Muhammad Umar
I need the answer now please
Anonymous
Look at David's question below this one.
Anonymous
then work done = (1/2) k x^2
use that k
x = 0.003 meters for answer in Joules
use that k
x = 0.003 meters for answer in Joules
henry2,
a. k = 100N/0.1cm = 1000N/cm.
*
b. F = 1000N/cm * 0.3cm = 300 N.
W = F*d = 300*0.003 = 0.9 Joules.
*
b. F = 1000N/cm * 0.3cm = 300 N.
W = F*d = 300*0.003 = 0.9 Joules.
No Henry !!!!
Henry, no, F not constant
Mistery
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
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
Cynthia
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
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
AIYEDUNSHO ENOCH
thank you the answer was perfect
Oyesijiayomide@gmail
4.0
Bot
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
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
Bot
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.