Question

An object of mass 100kg 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 1/40 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 10N-sec/m , find the equation of motion of the object. After how many seconds will the velocity of the object be 70 m/sec ?

Short answer

The result is $\mathrm{x} \left(\mathrm{t}\right)=95.65\mathrm{t}-956.5{\mathrm{e}}^{\frac{-\mathrm{t}}{10}}-956.5$ and time taken by the object is 13.2 sec .

Step-by-step solution

Important hint.

Use Newton’s method to solve for t ${\mathrm{t}}_{\mathrm{n}+1}={\mathrm{t}}_{\mathrm{n}}-\frac{\mathrm{f}({\mathrm{t}}_{\mathrm{n}})}{\mathrm{f}\text{'}({\mathrm{t}}_{\mathrm{n}})}$.

Find the equation of motion

The total force acting on the object is $\mathrm{F}-\mathrm{mg}-\mathrm{bv}-\frac{1}{40}\mathrm{mg}$.

Applying newton’s second law:

$100\frac{\mathrm{dv}}{\mathrm{dt}}=100 (9.81)-10 \mathrm{v}-\frac{1 \left(100\right) (9.81)}{40}$

$$\begin{gathered} \mathrm{v}\text{'}=9.56-0.1\mathrm{v} \\ \mathrm{v}\text{'}+0.1\mathrm{v}=9.56 \\ \mathrm{v}\text{'}+0.1\mathrm{v}=9.56 \left(\mathrm{onsolvingbyvariableseparating}\right) \end{gathered}$$

When $\mathrm{v} \left(0\right)=0, \mathrm{then} \mathrm{C}=-95.65$.

$$\begin{gathered} \mathrm{v}=95.65-95.65{\mathrm{e}}^{\frac{-\mathrm{t}}{10}} \\ \mathrm{x}\left(\mathrm{t}\right)=95.65\mathrm{t}-956.5{\mathrm{e}}^{\frac{-\mathrm{t}}{10}}+\mathrm{c} \end{gathered}$$ …… (1)

When $\mathrm{x}\left(0\right)=0, \mathrm{then} \mathrm{C}=-956.5$.

$\mathrm{x} \left(\mathrm{t}\right)=95.65 \mathrm{t}-956.5 {\mathrm{e}}^{\frac{-\mathrm{t}}{10}}-956.5$

Find the value of t

When the object travelling at the velocity 70m/sec then by equation (1).

$$\begin{gathered} 70=95.65 \mathrm{t}-956.5 {\mathrm{e}}^{\frac{-\mathrm{t}}{10}} \\ 70=95.65(1-{\mathrm{e}}^{\frac{-\mathrm{t}}{10}}) \\ \mathrm{t}=13.2 \sec \end{gathered}$$

Therefore, the result is $\mathrm{x} \left(\mathrm{t}\right)=95.65 \mathrm{t}-956.5 {\mathrm{e}}^{\frac{-\mathrm{t}}{10}}-956.5$ and $\mathrm{t}=13.2 \sec$.