Question

An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object​ down, a buoyancy force of StartFraction 1 Over 40 EndFraction times the weight of the object is pushing the object up​ (weight =​ mg). If we assume that water resistance exerts a force on the object that is proportional to the velocity of the​ object, with proportionality constant 10 ​N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 60 ​m/sec? Assume that the acceleration due to gravity is 9.81 m divided by sec squared.

Answer 1

Answer:

$v=+(9.56475-\frac{v}{8})\times t$

$t=57.9352\ s$

Explanation:

Given:

mass of the object, $m=80\ kg$

force of buoyancy acting on the object, $F_B=\frac{m.g}{40} =19.62\ N$

initial velocity of the object, $u=0\ m.s^{-1}$

Given that the water resistance acts on the object is proportional to $10\ N.s.m^{-1}$

So, the net force acting on the object:

$F=m.g-F_B+10\times v$

$F=784.8-19.62+10.v$

$F=765.18-10.v$ .........................................(1)

Now the acceleration of the object can be given as:

$a=\frac{F}{m}$

$a=\frac{765.18-10v}{80}$

$a=9.56475-\frac{v}{8}$

From the standard equation of motion:

$v=u+at$

$v=$ final velocity

$u=$ initial velocity

$t=$ time

$v=+(9.56475-\frac{v}{8})\times t$ .................................(2)

Now the velocity of the object is, $v=60\ m.s^{-1}$

$60=+(9.56475-\frac{60}{8})\times t$

$t=57.9352\ s$