The motion of a particle of mass m is described by the relation, $y = ut – \frac{1}{2}gt^2$, where $u$ is the initial velocity of the particle. The force acting on the particle is
[amp_mcq option1=”$F = m\left(\frac{du}{dt}\right)$” option2=”$F = mg$” option3=”$F = m\left(\frac{dy}{dt}\right)$” option4=”$F = -mg$” correct=”option4″]
This question was previously asked in
UPSC NDA-2 – 2023
The force acting on the particle is $F = -mg$.
The given equation of motion, $y = ut – \frac{1}{2}gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $\frac{dy}{dt} = u – gt$. The second derivative gives acceleration: $\frac{d^2y}{dt^2} = -g$. According to Newton’s second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. The negative sign indicates the force is acting in the negative y direction, which corresponds to the downward direction (gravity) if the initial upward direction was taken as positive y.