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As a part of an exercise routine to strengthen their pectoral muscles a person stretches a spring which has a spring constant k = 740 N/m. show answer No Attempt 50% Part (a) Calculate the work in joules required to stretch the springs from their relaxed state to the position x1 = 54 cm.

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Olajidey

Answer:

= 108J

Explanation:

To develop the problem it is necessary to apply the concepts related to Hooke's Law. For which it is understood that

[tex]F = kx[/tex]

Where,

k = Spring constant

x = Displacement of Spring

By the theorem of work we understand that it is the force realized when there is the path of an object therefore

[tex]dW =Fdx[/tex]

Our value of F is

[tex]dW = (kx)dx[/tex]

Integrating

[tex]W = k\int\limit_0^{x_1} xdx[/tex]

Therefore the expresion is

[tex]W = \frac{kx_1^2}{2}[/tex]

[tex]W = \frac{740 \times0.54^2 }{2}\\= 107.89J[/tex]

≅ 108J

asuwafoh

Answer:

107.892 J.

Explanation:

Work done to stretch a spring: This can be this can be defined as the amount of work needed to stretch a spring from its relaxed position. The S.I unit of work done is Joules (J).

The expression for work done to stretch a spring is

W = 1/2ke²............... Equation 1

Where k = spring constant, e = extension

Given: k = 740 N/m, e = 554 cm = 0.54 m

Substitute into equation 1

W = 1/2(740)(0.54²)

W = 370(0.2916)

W = 107.892 J.

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