Question

A 5.83 kg object is released from rest while fully submerged in a liquid. The liquid displaced by the submerged object has a mass of 2.72 kg. Take the upward direction to be positive. What is the displacement of the object in 0.143 s, assuming that it moves freely and that the drag force on it from the liquid is negligible?

Answer 1

displacement of the object is -005454 m

Step-by-step explanation:

Given data

mass m1 = 5.83 kg

mass m2 = 2.72 kg

time = 0.143 s

to find out

What is the displacement of the object

solution

we know the resultant force that is here

resultant force = m1(-g) + m2(g)

so m1 a = g ( m2 - m1 )

put all value and we get a

5.83 a = 9.81 ( 2.72 - 5.83)

a = - 0.5334 m/s²

and

now we get displacement from this formula

displacement = ut + 1/2 at²

here u = 0 and t = 0.143 and a = 0.5334 put all these

displacement = (0) (0.143) + 1/2 (-0.5334) (0.143)²

displacement = (0) (0.143) + 1/2 (-0.5334) (0.143)²

displacement = -005454 m

displacement of the object is -005454 m