Question

It takes 4.57 J of work to stretch a Hooke’s-law spring 9.11 cm from its unstressed length. How much the extra work is required to stretch it an additional 9.54 cm?Answer in units of J

Answer 1

The extra work required to stretch the spring an additional 9.54 cm can be calculated using the formula Ws = 1/2kx^2, where Ws is the work done, k is the spring constant, and x is the displacement from the unstressed length of the spring.

Step-by-step explanation:

The work done to stretch a spring can be calculated using the formula Ws = 1/2kx^2, where Ws is the work done, k is the spring constant, and x is the displacement from the unstressed length of the spring. In this case, it takes 4.57 J of work to stretch the spring 9.11 cm. To calculate the extra work required to stretch it an additional 9.54 cm, we need to find the new displacement by adding the initial displacement.

The new displacement is 9.11 cm + 9.54 cm = 18.65 cm. Plugging this value into the formula, we get:

Ws = 1/2k(18.65 cm)^2

We can solve this equation to find the extra work required to stretch the spring an additional 9.54 cm.

Answer 2

Step-by-step explanation:

Given that,

Work done, W = 4.57 J

Distance,
`x_1=9.11\ cm`

Distance,
`x_2=9.54\ cm`

We know that the work done by the spring is given by :

`W_=-kx_1^2`

`k=(W)/(x_1^2)`

`k=(4.57)/((9.11* 10^(-2))^2)`

k = 550.65 N/m

To find,

Let
`W_2` is the extra work is required to stretch it an additional 9.54 cm. It can be calculated as :

`W_2=(1)/(2)k(x_2^2-x_1^2)`

`W_2=(1)/(2)* 550.65(9.54^2-9.11^2)`

W = 2207.96 J

So, the extra wok done is 2207.96 J.