The magnitude of the external force acting on moving bodies can be known directly by

$A$ body of mass $0.05 \ kg$ is observed to fall with an acceleration of $9.5 \ ms^{-2}$. The opposing force of air on the body is $(g = 9.8 \ ms^{-2})$.

The velocity $(v)$ of a particle (under a force $F$) depends on its distance $(x)$ from the origin (with $x > 0$) as $v \propto \frac{1}{\sqrt{x}}$. Find how the magnitude of the force $(F)$ on the particle depends on $x$.

$A$ force vector applied on a mass is represented as $\vec{F} = 6\hat{i} - 8\hat{j} + 10\hat{k} \, N$ and it accelerates with $1 \, m/s^2$. What will be the mass of the body?

$A$ particle of mass $2/3 \, kg$ with velocity $v = -15 \, m/s$ at $t = -2 \, s$ is acted upon by a force $F = k - \beta t^2$. Here,$k = 8 \, N$ and $\beta = 2 \, N/s^2$. The motion is one-dimensional. Then,the speed at which the particle acceleration is zero again,is ........... $m/s$.

An object of mass $3\,kg$ is at rest. If a force $\vec{F} = (6t^2\hat{i} + 4t\hat{j})\,N$ is applied on the object,then the velocity of the object at $t = 3\,s$ will be