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Q5.2(b):
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (b) during its downward motion,

Solution :

Step 1: Identification of Forces Acting on the Pebble

When an object is in free fall or moving vertically, the net force ($F_{net}$) is defined by Newton's Second Law of Motion: $F = ma$. In the absence of air resistance, the only force acting on the pebble is the gravitational force ($F_g$), directed vertically downwards toward the center of the Earth. [Per Newton’s Law of Universal Gravitation].

Step 2: Analysis of Downward Motion

During the downward motion of the pebble, its velocity vector $\vec{v}$ is directed downwards. However, the force of gravity acts independently of the direction of motion or velocity. Gravity remains constant in magnitude and direction regardless of whether the pebble is moving upwards, downwards, or is momentarily at rest at the peak of its trajectory.

Step 3: Mathematical Calculation

Given the mass of the pebble $m = 0.05 \text{ kg}$ and the acceleration due to gravity $g \approx 9.8 \text{ m/s}^2$:

$F_{net} = m \cdot g$

$F_{net} = 0.05 \text{ kg} \times 9.8 \text{ m/s}^2$

$F_{net} = 0.49 \text{ N}$

Step 4: Vector Representation

The direction of the force is always towards the Earth (vertically downwards). Even during downward motion, the acceleration remains downwards ($a = -g$), causing the pebble to accelerate at $9.8 \text{ m/s}^2$ in the direction of its motion.

Pebble F = 0.49 N Direction of Motion (Downward)

Summary of Dynamics

Parameter Value/Direction
Mass ($m$) $0.05 \text{ kg}$
Acceleration ($g$) $9.8 \text{ m/s}^2$
Magnitude of Force $0.49 \text{ N}$
Direction of Force Vertically Downward

Final Solution: The net force on the pebble during its downward motion is 0.49 N, directed vertically downwards.


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