Hooke's Law

Classical Mechanics Level 2

A spring is stretched by 5.0 cm when a force of 65.0 N is applied to it. What is its spring constant in N/m? Answer to the correct number of significant figures.

Hooke's Law: F s = − k x F_{s}=-kx


The answer is 1.3E+3.

Sir Constructalot by Sir Constructalot , 21, Canada

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2 solutions

Mahdi Raza
Jun 10, 2020

5 5 cm = 0.05 0.05 m, substitute that as x x in the formula and we get:

65 = ( k ) ( 0.05 ) k = 65 0.05 k = 1300 \begin{aligned} 65 &= (k)(0.05) \\ k &= \dfrac{65}{0.05} \\k &= \boxed{1300} \end{aligned}

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Sir Constructalot
Jun 10, 2020

Question: A spring is stretched by 5.0 cm when a force of 65.0 N is applied to it. What is its spring constant in N/m? Answer to the correct number of significant figures.

To solve this problem, use Hooke's Law. First, the 5.0 cm must be converted to metres, then you can manipulate the equation to solve for k, the spring constant.

  1. Convert 5.0 cm to m

    5.0cm x 1 m 100 c m \frac{1m}{100cm} = 0.050m

  2. Manipulate the equation and solve.

    F s = k x F_{s} = kx

    k = F s x \frac{F_{s}}{x}

    k= 65.0 N 0.050 m \frac{65.0N}{0.050m}

    k= 1300 N/m

    k= 1.3E3 N/m

1 Helpful 0 Interesting 0 Brilliant 0 Confused

The Hooke's law was F s = − k x F_s=-kx not F s = k x F_s=kx

Zakir Husain - 1 year ago

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The magnitude of the force is k x kx . The law is just written with the negative sign to show that the force of the spring is acting opposite to the direction of stretch. If it is stretch a distance of x x , then the force that is required to stretch it to that displacement is k x kx .

Krishna Karthik - 12 months ago

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