Stretched 1 m beyond its natural length, a certain spring exerts a restoring force of magnitude 50 newtons. How much work, W, does it take to stretch the spring 7 m beyond its natural length?

Respuesta :

Answer:

1,225 J

Step-by-step explanation:

The restoring force exerted by a spring is given by:

[tex]F=kx[/tex]

Where 'k' is the spring constant and 'x' is the stretched distance.

Since x=1, the spring constant is:

[tex]k=\frac{50}{1}= 50\ N/m[/tex]

The work required to stretch a spring is given by:

[tex]W=\frac{kx^2}{2}[/tex]

The work needed to stretch this spring 7 m beyond its natural length is:

[tex]W=\frac{50*(7^2)}{2}\\W=1,225\ J[/tex]

The work required is 1,225 J.

The amount of workdone to stretch the spring 7 m beyond its natural length is 1225 Joules.

The restoring force exerted by a spring is given by:

[tex]F=kx[/tex]

[tex]k=\frac{F}{x} \\k=\frac{50N}{1m}\\k=50N/m[/tex]

The formula for calculating the workdone is expressed as:

[tex]W=\frac{1}{2}kx^2[/tex]

Substitute the given parameters;

[tex]W=\frac{1}{2}(50)\times7^2\\W=25 \times 49\\W= 1,225Joules[/tex]

Hence the amount of workdone to stretch the spring 7 m beyond its natural length is 1225 Joules.

Learn more on workdone here: https://brainly.com/question/9821607

ACCESS MORE