Answer :
a) If the spring stretch's from 42cm and 46 cm is 2.78J
b) 3.64cm length far beyond will stretch from 30cm.
Natural Length:
The natural length of a spring is the length with no mass attached. Assume that the spring obeys Hooke's law. When the spring length changes from its natural length by ΔL, the spring exerts a force of Fs=kΔL. where k is a positive number, denoted as the spring constant.
Natural length is 36 and stretches to 45. 5 J of work is needed
⇒ 5 = 1/2k(0.45-0.36)²
⇒ 5 = 1/2k(0.09)²
⇒ 5 = 0.0081k/2
⇒ 10 = 0.0081k
⇒ K = 1234.568
Now,
X1 = 0.38-0.36 = 0.02
X2 = 0.43-0.36 = 0.07
Work done
W = 1/2k(x2²-x1²)
W = 1/2(1234.568)(0.007²-0.02²)
W = 1234.568(0.0049-0.0004)/2
W = 2.78j
F = 45N
F = kx
X = f/k
= 45/1234.568
= 0.0364
= 3.64cm
The answer to A is 2.78j
The answer to b is 3.64cm
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