Problem 400 - JEE Physics

Question

The motion of a particle of a mass \(m\) is describe by \(\mathrm{y}=\mathrm{ut}+(1 / 2) \mathrm{gt}^{2}\). Find the force acting on the particle. (A) \(\mathrm{F}=\mathrm{ma}\) (B) \(\mathrm{F}=\mathrm{mg} \quad\) (C) \(\mathrm{F}=0\) (D) None of these

Step-by-Step Solution

Answer

The force acting on the particle is given by Newton's second law, F = ma. The acceleration (a) of the particle is equal to the gravitational acceleration (g), which is derived from the given equation of motion. Therefore, the force acting on the particle is F = mg, which corresponds to option (B).

Step 1: Identify the acceleration from the given equation

The given equation is y = ut + (1/2)gt^2, where u is the initial velocity, g is the gravitational acceleration, and t is time. The acceleration, in this case, is the second derivative of the equation concerning the time.

Step 2: Differentiate the equation concerning time twice

First, find the velocity (the first derivative of y with respect to time): v = dy/dt = u + gt Now, find the acceleration (the second derivative of y with respect to time): a = dv/dt = d(u + gt)/dt = 0 + g = g The acceleration (a) of the particle is g.

Step 3: Use Newton's second law to find the force acting on the particle

Newton's second law states that the force acting on an object is equal to the product of mass and acceleration (F = ma). Using the mass (m) and the found acceleration (g): F = m * a = m * g

Step 4: Match the found force with the given options

We found that the force acting on the particle is F = mg. This matches option (B). Therefore, the correct answer is (B) F = mg.

Key Concepts

Motion Equations

Motion equations are fundamental in physics as they describe the movement of objects through space and time. In the given exercise, we see that the motion is described by the equation:

• \( y = ut + \frac{1}{2}gt^2 \)

Here, \( y \) represents the displacement of the particle, \( u \) is the initial velocity, \( g \) is the gravitational acceleration, and \( t \) is the time elapsed.
Motion equations help us predict the future position of an object and understand how various forces, like gravity, influence its path. They are derived by integrating acceleration equations over time, considering initial conditions of motion.
In practice, these equations allow us to calculate:

• Displacement over a time interval
• The velocity of an object at a given point in time
• Potential force interactions if combined with Newton's laws

Understanding how to manipulate and apply these equations is crucial for solving real-world motion problems.

Gravitational Acceleration

Gravitational acceleration, represented by the symbol \( g \), is a constant that quantifies the acceleration of an object due to the force of gravity.
On Earth, this acceleration is approximately \( 9.8 \, m/s^2 \), though it can differ slightly based on location. The given motion equation incorporates \( g \) to account for the effect of gravity on the particle as it moves.
It's important to note:

• \( g \) is always directed towards the center of the Earth, often considered negative when upwards is positive.
• Gravitational acceleration contributes to the object increasing or decreasing its speed when moving under the influence of gravity alone.

In our exercise, substituting \( g \) into the motion equation enables us to determine how quickly the particle accelerates downwards as time progresses.
This acceleration is crucial for solving problems involving free fall or the vertical motion of objects.

Differentiation in Physics

Differentiation is a powerful mathematical tool used to find rates of change, a key concept in motion analysis.
In our exercise, we use differentiation to find the velocity and acceleration of the particle from the displacement equation.
Here's how it works:

• First, we differentiate the displacement \( y \) with respect to time \( t \) to find velocity \( v \):
  \( v = \frac{dy}{dt} = u + gt \)
• Then, differentiate the velocity \( v \) with respect to \( t \) to find acceleration \( a \):
  \( a = \frac{dv}{dt} = g \)

Differentiation allows us to go from position to velocity to acceleration, which is foundational in understanding how an object's motion evolves.
In physics, mastering differentiation helps interpret changing physical quantities and analyze motion extensively.