A 1.5 kg rock is released from rest at the surface of a calm lake. If the resistance offered by the water as the rock falls is directly proportional to the rock’s velocity, the rock’s acceleration is a = g - Cd v/m, where g is the acceleration of gravity, Cd is a constant drag coefficient, v is

Question

A 1.5 kg rock is released from rest at the surface of a calm lake. If the resistance offered by the water as the rock falls is directly proportional to the rock’s velocity, the rock’s acceleration is [latex]a=g-C_d v / m[/latex], where g is the acceleration of gravity, [latex]C_d[/latex] is a constant drag coefficient, [latex]v[/latex] is the rock’s velocity, and m is the rock’s mass. Letting [latex]C_d[/latex] = 4.1 kg/s, determine the rock’s velocity after 1.8 s.

Accepted Answer

We begin by observing tha

[latex]\begin{aligned}a=g-\frac{C_d}{m} v=\frac{d v}{d t} & \Rightarrow \int_0^{t_f} d t=\int_0^{v_f} \frac{d v}{g-\left(C_d / m ight) v} \& \Rightarrow \quad t_f=\frac{-m}{C_d} \ln \left(\frac{g-\left(C_d / m ight) v_f}{g} ight) \quad \Rightarrow \quad t_f=\frac{-m}{C_d} \ln \left(1-\frac{C_d}{m g} v_f ight), ext {(1)}\end{aligned}[/latex]

where the subscript f stands for final. Solving the above equation for [latex]v_f[/latex] , we have

[latex]v_f=\frac{m g}{C_d}\left(1-e^{-C_d t_f / m} ight) \Rightarrow v_f=3.56 \ m / s[/latex]

where we have used the following numerical data: m = 1.5 kg, g = 9.81 m/s² , [latex]C_d[/latex] = 4.1 kg/s, and [latex]t_f[/latex] = 1.8 s.

Question 2.50: A 1.5 kg rock is released from rest at the surface of a calm...

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