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.
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