A water-filled beaker has a spring with spring constant, k = 130 N/m, attached t

Question

A water-filled beaker has a spring with
spring constant, k = 130 N/m, attached
to its bottom as shown in the figure.

When a wooden block of mass m = 4.20 kg
and density = 630 kg/m3 is connected to the
spring, it stretches upward a distance L as shown.
Use the variables listed in the problem
m, k, , L along with g and
w (density of water).

(a) Write a symbolic force equation for the block, assuming that it has reached equilibrium (and taking up as positive).

=  =

(b) How far did the spring stretch?

L = m

Explanation / Answer

here,

Assuming :
Fb = Bouyant force
Fs = Spring Force
Fg = force due to gravity

spring constant = 130 N/m

mass of wodden block, mb = 4.20kg
density of block, pb = 630 kg/m^3
density of water, pw = 1000 kg/m^3

Voulme of block, Vb
Vb = mb/pb
Vb = 4.20/630
Vb = 6.6 * 10^-3 m^3

From new ton second Law, SUM(F) = 0
Fb - Fs = 0
Fb = Fs

Pw*Vb*g - mb*g = K*del(l)

Mb * g * (1 - (pw/pb) ) = K*del(l)

Part B :

Bouyancy force :
Fb = Mb * g * (1 - (pw/pb) )
Fb = 4.20 * 9.8 * (1 - (1000/630) )
Fb = - 24.173 N

this is upward pull on the spring>>>
restoring force = + 24.173 N downwards

Spring Force = k*del(l)
24.173 = 130 * del(l)
Del(l) = 0.185 m

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