Question

Calculate the work required to stretch the following springs 0.60.6 m from their equilibrium positions. Assume​ Hooke's law is obeyed. a. A spring that required a force of 4040 N to be stretched 0.10.1 m from its equilibrium position. b. A spring that required 4040 J of work to be stretched 0.10.1 m from its equilibrium position.

Answer 1

Step-by-step explanation:

Given that,

Distance = 0.6 m

Force = 40 N

(a). We need to calculate the spring constant

Using Hooke's law

`F=kx`

Put the value into the formula

`40=k*0.1`

`k=(40)/(0.1)`

`k=400`

We need to calculate the work done

`W=\int_(0)^(0.6){kx}dx`

`W=\int_(0)^(0.6){400x}dx`

On integrating

`W=400*{(x^2)/(2)}_(0)^(0.6)`

`W=400{(0.6^2)/(2)-0}`

`W=72\ J`

(b). We need to calculate the spring constant

Using formula of work done

`W=\int_(0)^(0.1){kx}dx`

`40=\int_(0)^(0.1){kx}dx`

`40=k((x^2)/(2))_(0)^(0.1)`

`40=k*{(0.1^2)/(2)-0}`

`40=k*0.005`

`k =(40)/(0.005)`

`k=8000`

We need to calculate the work done

`W=\int_(0)^(0.6){k x}dx`

`W=8000*((x^2)/(2))_(0)^(0.6)`

`W=8000*(0.6^2)/(2)-0}`

`W=1440\ J`

Hence, This is the required solution.