Physics Solution: Object Deceleration in Viscous Liquid

This solution calculates the time taken for an object to decelerate in a viscous liquid under a specific drag force, focusing on the key physics principles involved.

Problem Parameters

• Object mass: $m = 1 \text{ kg}$
• Initial speed: $v_o$
• Drag force: $D = \alpha v$, where $\alpha = 1 \text{ kg/s}$
• Gravity: Ignored
• Target speed: $v_o/2$

Applying Newton's Second Law

The net force acting on the object is the resistive drag force, which opposes motion. According to Newton's second law ($F_{net} = ma$), the equation of motion is:

$ m a = -D $

Substituting the given drag force form ($D = \alpha v$) and acceleration ($a = \frac{dv}{dt}$):

$ m \frac{dv}{dt} = -\alpha v $

Solving the Differential Equation for Time

To determine the time ($t$), we rearrange the equation into a separable differential form:

$ \frac{dv}{v} = -\frac{\alpha}{m} dt $

Now, we integrate both sides. The speed decreases from the initial speed $v_o$ to the final speed $v_o/2$ over the time interval from $t=0$ to $t$.

$ \int_{v_o}^{v_o/2} \frac{dv}{v} = \int_{0}^{t} -\frac{\alpha}{m} dt $

Evaluating the integrals:

$ [\ln|v|]_{v_o}^{v_o/2} = -\frac{\alpha}{m} [t]_{0}^{t} $

$ \ln(v_o/2) - \ln(v_o) = -\frac{\alpha}{m} t $

Using the logarithm property $\ln(a) - \ln(b) = \ln(a/b)$:

$ \ln\left(\frac{v_o/2}{v_o}\right) = -\frac{\alpha}{m} t $

$ \ln(1/2) = -\frac{\alpha}{m} t $

Since $\ln(1/2) = -\ln(2)$:

$ -\ln(2) = -\frac{\alpha}{m} t $

Solving for $t$ gives the formula for the time taken:

$ t = \frac{m}{\alpha} \ln(2) $

Calculating the Specific Time

Substitute the given values $m = 1 \text{ kg}$ and $\alpha = 1 \text{ kg/s}$ into the derived formula:

$ t = \frac{1 \text{ kg}}{1 \text{ kg/s}} \ln(2) $

$ t = 1 \times \ln(2) \text{ s} $

Using the approximate value for the natural logarithm of 2, $\ln(2) \approx 0.693147$:

$ t \approx 0.693 \text{ s} $

This calculated time of approximately 0.693 seconds falls within the expected range of 0.67 to 0.71 seconds.