Asked by Muhammad Umar
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
Answered by
Muhammad Umar
I need the answer now please
Answered by
Anonymous
Look at David's question below this one.
Answered by
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
Answered by
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.
Answered by
No Henry !!!!
Henry, no, F not constant
Answered by
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
Answered by
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
Answered by
AIYEDUNSHO ENOCH
thank you the answer was perfect
Answered by
Oyesijiayomide@gmail
4.0
Answer
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
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