An object having a mass of 5.0 kg is placed beneath the surface of a lake. A buo

Question

An object having a mass of 5.0 kg is placed beneath the surface of a lake. A buoyant force of magnitude 53.0 N acts on the object, causing it to rise. The water exerts a resistance force whose magnitude is numerically equal to the square of the instantaneous velocity of the object. Five seconds after it starts to rise, the object reaches the surface of the lake. a) Take the direction of your coordinate axis to point upward. Write the differential equation and initial condition for the velocity v(t) of the object as a function of time. b) Find the velocity of the object at the instant it reaches the surface (take the value of the gravitational acceleration constant g to be 9.8 m/s^2

Explanation / Answer

here mg = 5 * 9.8 = 49 N Fb ( buoyant force ) = 53 N F ( resistance ) = v^2 v: instantaneous velocity of object Form force balance equation Fb - mg - F = m*a a : instantaneous acceleration of object = dv / dt on putting values 53 - 49 - v^2 = 5 * ( dv / dt ) a) Differential equation : dv / dt = ( 4 - v^2 ) / 5 b) dv / ( 4 - v^2 ) = dt / 5 on integrating since ? dy / a^2 - y^2 = ln ( a + x/ a - x ) } / 2a , here a = 2 limit for v from zero to v for t from zero to 5 ln ( 2 + v / 2 -v) } / 4 = t / 5 on applying limits v = 2 * ( e^4 -1 ) / ( e^4 + 1) v = 1.92 m/s

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