Suppose that 5 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 45 cm. Required:
a. How much work is needed to stretch the spring from 38 cm to 43 cm? (Round your answer to two decimal places.) J
b. How far beyond its natural length will a force of 45 N keep the spring stretched? (Round your answer one decimal place.

2.78J

3.64cm

Explanation:

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

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