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You find that if you hang a 1.25kg weight from a vertical spring, it stretches 3.75cm.

a) What is the force constant of this spring?

b) How much mass should hang from this spring so it will stretch 8.13 cm from its original unstretched length?

Answer :

akande212

Answer:

(A) k = 327N/m

(B) m = 2.71kg

Explanation:

The weight of the 1.25kg is the force acting on the spring. W= weight = mg =F = kx

That is

mg = kx

Therefore k = mg/x

Given m = 1.25kg and x = 3.75cm = 0.0375m

k = 1.25 ×9.8 / 0.0375 = 327N/m

(B) from the relation mg =kx

Given x = 8.13cm = 0.0813m.

m = kx / g = 327 × 0.0813 / 9.8 = 2.17kg.

ajeigbeibraheem

The force constant (k) of the spring is 326.67 N/m. The amount of mass that should be hung from this spring is 2.71 kg

From the information given;

  • the weight mass = 1.25 kg
  • the distance(x) with which the object stretches = 3.75 cm

The force constant of the spring according to Hooke law of elasticity can be expressed as:

F = kx

where;

  • F = mg  

∴

mg = kx

Making the force constant k the subject of the formula, then:

[tex]\mathbf{k = \dfrac{mg}{x}}[/tex]

[tex]\mathbf{k = \dfrac{1.25 \times 9.8}{3.75 }}[/tex]

[tex]\mathbf{k =326.67 N/m}[/tex]

However, if the distance(x) is stretched at 8.13 cm, the mass at which the object should be hung from this spring is determined by using the formula;

  • mg = kx

Making mass (m) the subject, we have:

[tex]\mathbf{m = \dfrac{k \times( x)}{g}}[/tex]

[tex]\mathbf{m = \dfrac{326.67 \times0.0813}{9.8}}[/tex]

m = 2.71 kg

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