Respuesta :
The force of a spring can be calculated by the expression:
F = kx where k is a constant and x is the distance.
From the problem statement, we can solve the value of k from the initial conditions given as F = 25 N and x = 2 cm.
25 = k (0.02)
k = 1250
We use the work formula which is force times the distance resulting to the expression:
W = (1/2)(kx^2)
W = (1/2)(1250)(0.03^2)
W = 0.5625 J
F = kx where k is a constant and x is the distance.
From the problem statement, we can solve the value of k from the initial conditions given as F = 25 N and x = 2 cm.
25 = k (0.02)
k = 1250
We use the work formula which is force times the distance resulting to the expression:
W = (1/2)(kx^2)
W = (1/2)(1250)(0.03^2)
W = 0.5625 J
Answer: The work done in further extending spring by 3 cm is 0.5625 Joules.
Explanation:
Force exerted on spring,F = 25 N
Distance up to which spring stretched ,x= 2 cm = 0.02 m
(1 m = 100 cm)
[tex]F=kx[/tex]
[tex]k=\frac{F}{x}=\frac{25 N}{0.02 m}=1250 N/m[/tex]
Work done while further stretching the spring by distance x' = 3 cm.
Work done = [tex]\frac{1}{2}k\times x'^2=\frac{1}{2}\times 1250 N/m\times (3\times 10^{-2} m)^2=0.5625 N m=0.5625 Joules[/tex]
The work done in further extending spring by 3 cm is 0.5625 Joules.