Answer to Question #298333 in Mechanics | Relativity for MishaTA

Question

A particle of mass 3 Kg falls from rest at a point 5 m above the surface of a liquid which is in a container. There is no instantaneous change in speed of the particle as it enters the liquid. The depth of the liquid in the container is 4 m. The downward acceleration of the particle while it moving in the liquid is 5.5 m/s^2. (i) Find the resistance to motion of the particle while it is moving in the liquid. (ii) Sketch the velocity- time graph for the motion of the particle, from the time it starts to move until the time it reaches the bottom of the container. Show on your sketch the velocity and the time when the particle enters the liquid, and when the particle reaches the bottom of the container.

Answer 1

Explanations & Calculations

a)

• Apply $\small F= ma$ on the object for its downward motion.

$\qquad\qquad \begin{aligned} \small mg-f&=\small ma\\ \small f&=\small m(g-a) \end{aligned}$

b)

• Since the net force imposed on the object that is $\small \downarrow mg-f$ is constant throughout, it experiences a constant acceleration.
• Therefore, the v-t graph is just a straight line ($\small y=mx+c$ ) starting at $\small v_{initial} = \sqrt{2gh}=\sqrt{2g\times5}$ and extending away from there at a gradient equal to the magnitude of the acceleration $\small m= 5.5$ .