Question

A light spring of constant 176 N/m rests vertically on the bottom of a large beaker of water.
A 4.63 kg block of wood of density 648 kg/m^3
is connected to the top of the spring and
the block-spring system is allowed to come to
static equilibrium.
What is the elongation ∆L of the spring?
The acceleration of gravity is 9.8 m/s^2
Answer in units of cm.

Answer 1

Answer:

14.0 cm

Explanation:

Draw a free body diagram of the block.  There are three forces: weight force mg pulling down, elastic force k∆L pulling down, and buoyancy ρVg pushing up.

Sum of forces in the y direction:

∑F = ma

ρVg − mg − k∆L = 0

(1000 kg/m³) (4.63 kg / 648 kg/m³) (9.8 m/s²) − (4.63 kg) (9.8 m/s²) − (176 N/m) ∆L = 0

∆L = 0.140 m

∆L = 14.0 cm

Answer 2

We have that for the Question "A light spring of constant 176 N/m rests vertically on the bottom of a large beaker of water." it can be said that

x=0.19m

From the question we are told

A light spring of constant 176 N/m rests vertically on the bottom of a large beaker of water.

• A 4.63 kg block of wood of density 648 kg/m^3  is connected to the top of the spring and  the block-spring system is allowed to come to  static equilibrium.

• What is the elongation ∆L of the spring?
• The acceleration of gravity is 9.8 m/s^2

Generally the equation for the Bouyancy force  is mathematically given as

$F= mg*(\frac{1-pH2O}{p}\\\\F=5.5*9.81*(\frac{1-1000}{632}\\\\F=-31.4$

Therefore

$kx=F\\\\x=\frac{31.4}{166}$

x=0.19m

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