Answer to Question #133366 in Mechanics | Relativity for Emma

Question

You are on a fishing trip, not catching anything, and bored - time for physics! You drop your 50g weight to the bottom of the lake under constant tension from the string of magnitude 0.1N. The damping constant for your falling weight is 2.5kg/s.

a) What is the terminal velocity of the weight? Give your answer in m/s and assume the positive direction is upwards.

b) If it takes 500s for the weight to reach the bottom of the lake, how deep is the lake? Give your answer in m.

Answer 1

Explanations & Calculations

• $\small m = 0.05kg$ & damping constant = $\small 2.5\times 9.8 Ns^{-1}$
• At the terminal velocity, forces acting on the object should be equal, therefore,

$\qquad\qquad \begin{aligned} \small mg&=\small t+f\\ \small 0.5 &= \small 0.1+f\\ \small f &= \small 0.4N \end{aligned}$

• Now if we consider the damping force on the object to be $\small f$ , then with the damping constant it's possible to write $\small f=f_o+(2.5\times 9.8)t$. Since there is no any drag force at the beginning at $\small t=0$ ,then $\small f=24.5t$ .
• Then time to achieve terminal velocity could be calculated as,

$\qquad\qquad \begin{aligned} \small 24.5t= \small 0.4 \to t = 0.01633s \end{aligned}$

• Then to calculate the terminal velocity, consider a position at time $\small t$ before achieving the terminal velocity & apply Newton's second law along the moving direction $\big \downarrow$

$\qquad\qquad \begin{aligned} \small mg - T-f &= \small m a\\ \small 0.5-0.1-25t&= \small 0.05\frac{dv}{dt}\\ \small \int_{0}^{v} dv &= \small \frac{1}{0.05}\int_{0}^{t}\Big(0.4-25t \Big)dt\\ \small v &= \small \frac{1}{0.05} \Big(0.4t-12.5t^2 \Big)|_{0}^{t}\\ \small v_t &= -\small \bold{0.6397ms^{-1}} \end{aligned}$ : $\small a$ is not constant

• Time spent at the terminal velocity = 500 - 0.01633 = 499.9837s
• Time taken to achieve terminal velocity could be neglected, then its the motion at a uniform velocity, therefore,

$\qquad\qquad \begin{aligned} \end{aligned}$$\qquad\qquad \begin{aligned} \small S &= ut\\ \small h &= -\small 0.6397ms^{-1} \times 500 s\\ &= -\small \bold{319.85m} \end{aligned}$