Suppose that 5 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 39 cm.

Question

Suppose that 5 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 39 cm

.

(a) How much work is needed to stretch the spring from 33 cm to 35 cm? (Round your answer to two decimal places.)

 _____J

(b) How far beyond its natural length will a force of 30 N keep the spring stretched? (Round your answer one decimal place.)

_____ cm

Answer 1

I will answer this question as if it were a physics question.

I am not sure why it is in a calculus class.

A spring has a spring constant k with the following properties

1) To stretch the spring extra length x requires force kx

2) A spring stretched length x has energy kx²/2.

The given information allows you to calculate k as follows.

Work of 5J gave the spring energy k(39-28)²/2.

Therefore

5 = k(39-28)²/2

k = 10/121 N/cm

(a)

At 33cm the spring has energy k(33-28)²/2.

At 35cm the spring has energy k(35-28)²/2.

The extra energy needed is

(10/121)(35-28)²/2 - (10/121)(33-28)²/2 = 120/121 = 0.99J

(b)

30 = kx

x = 30/k = 363cm