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A person who weighs 685 N steps onto a spring scale in the bathroom, and the spring compresses by 0.88 cm. (a) What is the spring constant? N/m (b) What is the weight of another person who compresses the spring by 0.38 cm?

Answer :

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To solve this problem it is necessary to use the concepts of Force of a spring through Hooke's law, therefore,

[tex]F = kx[/tex]

Where,

k = Spring constant

x = Displacement

Initially our values are given,

[tex]F = 685N[/tex]

[tex]x = 0.88 cm[/tex]

PART A ) With this values we can calculate the spring constant rearranging the previous equation,

[tex]k = \frac{F}{x}[/tex]

[tex]k = \frac{685}{0.88*10^-2}[/tex]

[tex]k = 77840.9N/m[/tex]

PART B) Since the constant is unique to the spring, we can now calculate the force through the new elongation (0.38cm), that is

[tex]F = kx[/tex]

[tex]F = (77840.9)(0.0038)[/tex]

[tex]F = 295.79N[/tex]

Therefore the weight of another person is 265.79N.

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