Question

It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of the spring constant of this spring?

Answer 1

Answer:

43.56 N/m

Explanation:

From Hook's Law,

The Energy stored in a spring is given as

E = 1/2ke².......................... Equation 1

Where E = Energy stored in the spring, k = spring constant of the spring, e =. extension

make k the subject of the equation

k = 2E/e²....................... Equation 2

Given: E = 49 J, e = 2.9-1.4 = 1.5 m.

Substitute into equation 2

k = 2(49)/1.5²

k = 98/2.25

k = 43.56 N/m.

Hence the spring constant of the spring = 43.56 N/m

Answer 2

Answer:

44 N/m

Explanation:

The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m

The work needed to stretch a spring by e is given by

$W = \frac{1}{2} ke^2$

where k is spring constant.

$k = \dfrac{2W}{e^2}$

Using the appropriate values,

$k = \dfrac{2\times 49\text{ J}}{1.5^2\text{ m}^2} = 43.55\ldots\text{ N/m} \approx 44\text{ N/m}$