Understanding the Particle Motion and Finding Force

The question describes the motion of a particle of mass \(m\) using the equation \(y = ut - \frac{1}{2} gt^2\). This equation is a standard kinematic equation describing the vertical displacement (\(y\)) of an object under constant acceleration, which in this case is the acceleration due to gravity (\(g\)). Here, \(u\) represents the initial velocity of the particle, and \(t\) is time.

To find the force acting on the particle, we need to determine its acceleration. According to Newton's Second Law of Motion, the force (\(F\)) acting on an object is equal to its mass (\(m\)) multiplied by its acceleration (\(a\)):

\[F = ma\]

The acceleration is the rate of change of velocity with respect to time, and velocity is the rate of change of displacement with respect to time. We can find the acceleration by taking the second derivative of the displacement equation \(y\) with respect to time \(t\).

Step-by-Step Calculation of Force

Let's find the velocity (\(v\)) first by differentiating the displacement \(y\) with respect to \(t\):

Given displacement: \[y = ut - \frac{1}{2} gt^2\]

Velocity is the first derivative of displacement:

\[v = \frac{dy}{dt}\]

Differentiating \(y\) with respect to \(t\):

\[v = \frac{d}{dt}(ut - \frac{1}{2} gt^2)\]

Using the rules of differentiation (\(\frac{d}{dt}(ct) = c\) and \(\frac{d}{dt}(ct^n) = nct^{n-1}\)):

\[v = u \cdot \frac{d}{dt}(t) - \frac{1}{2} g \cdot \frac{d}{dt}(t^2)\]

\[v = u \cdot 1 - \frac{1}{2} g \cdot (2t)\]

\[v = u - gt\]

Now, let's find the acceleration (\(a\)) by differentiating the velocity \(v\) with respect to \(t\):

Acceleration is the first derivative of velocity:

\[a = \frac{dv}{dt}\]

Differentiating \(v\) with respect to \(t\):

\[a = \frac{d}{dt}(u - gt)\]

Using the rules of differentiation (\(\frac{d}{dt}(c) = 0\) for a constant \(c\), and \(\frac{d}{dt}(ct) = c\)):

\[a = \frac{d}{dt}(u) - \frac{d}{dt}(gt)\]

Assuming \(u\) is the initial velocity and is constant over time, \(\frac{du}{dt} = 0\). Note that $g$ is also a constant (magnitude of acceleration due to gravity).

\[a = 0 - g \cdot \frac{d}{dt}(t)\]

\[a = -g \cdot 1\]

\[a = -g\]

So, the acceleration of the particle is \(-g\). The negative sign indicates that the acceleration is in the direction opposite to the positive \(y\) direction (assuming positive \(y\) is upwards, gravity acts downwards).

Now, we can find the force acting on the particle using Newton's Second Law, \(F = ma\):

\[F = m(-g)\]

\[F = -mg\]

This force is the gravitational force acting on the particle, directed downwards.

Comparing with Given Options

Let's compare our result \(F = -mg\) with the provided options:

• Option 1: \(F = m (\frac{du}{dt})\). Since \(u\) is initial velocity, it's usually constant, so \(\frac{du}{dt} = 0\), giving \(F = 0\). This is incorrect.
• Option 2: F = mg. This is the magnitude of the force, but the equation \(y = ut - \frac{1}{2} gt^2\) implies the acceleration is \(-g\), leading to a force of \(-mg\).
• Option 3: \(F = m (\frac{dy}{dt})\). \(\frac{dy}{dt}\) is velocity (\(v\)), not acceleration (\(a\)). So this is \(F = mv\), which is incorrect for force under constant acceleration.
• Option 4: F = - mg. This matches our derived force.

Based on the analysis, the force acting on the particle is \(F = -mg\).

Revision Table: Key Concepts

Additional Information on Kinematics and Force

The equation \(y = ut - \frac{1}{2} gt^2\) is derived assuming that the acceleration is constant and equal to \(-g\), where the positive direction is upwards. This describes projectile motion in the vertical direction under the influence of gravity alone (neglecting air resistance). The initial velocity \(u\) is the initial vertical velocity.

• Kinematic equations relate displacement, velocity, acceleration, and time under constant acceleration. The general forms are:
  • \(v = u + at\)
  • \(y = ut + \frac{1}{2} at^2\) (or \(s = ut + \frac{1}{2} at^2\))
  • \(v^2 = u^2 + 2ay\) (or \(v^2 = u^2 + 2as\))
• Comparing the given equation \(y = ut - \frac{1}{2} gt^2\) with \(y = ut + \frac{1}{2} at^2\), we can directly see that the acceleration \(a\) is equal to \(-g\).
• A force of \(-mg\) is the gravitational force acting on a mass \(m\) near the Earth's surface, assuming positive is upwards. The magnitude is \(mg\), and the direction is downwards.

This problem highlights the connection between the description of motion (kinematics) and the cause of motion (dynamics, specifically forces). By analyzing the kinematic equation, we can deduce the acceleration, and subsequently, the force responsible for that acceleration.