Q.A pebble of mass $0.05\ \text{kg}$ is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

The key idea is Newton's Second Law, which states that the net force on an object is equal to its mass times its acceleration ($\vec{F}_{\text{net}} = m\vec{a}$). In the absence of air resistance, the only force acting on the pebble during its flight is gravity.

1. The mass of the pebble is $m = 0.05\ \text{kg}$, and the acceleration due to gravity is $g = 9.8\ \text{m/s}^2$.
2. Ignoring air resistance, the only force acting on the pebble at any point after it leaves the hand is the gravitational force.
3. This gravitational force is $\vec{F}_g = m\vec{g}$, which is always directed vertically downwards.
4. Therefore, the net force on the pebble is constant in both magnitude and direction throughout its motion, regardless of its velocity or position.

The magnitude of the net force is $F_{\text{net}} = mg = 0.05\ \text{kg} \times 9.8\ \text{m/s}^2 = 0.49\ \text{N}$.

The answers do not change if the pebble is thrown at an angle of $45^{\circ}$ with the horizontal. The gravitational force remains the only force acting on the pebble, and its magnitude and direction are independent of the projection angle.

When we talk about the "net force" on an object, we are directly referring to Newton's Second Law of Motion. This fundamental law states that the net force acting on an object is equal to the product of its mass and its acceleration. Mathematically, this is expressed as $\vec{F}_{net} = m\vec{a}$. The direction of the net force is always the same as the direction of the acceleration.

In this problem, we are asked to ignore air resistance. This is a crucial simplification because it means the only force acting on the pebble throughout its entire flight, once it leaves the hand, is the force of gravity. Gravity always acts vertically downwards, pulling the object towards the center of the Earth.

Let's calculate the magnitude of this gravitational force first.

Given:

Mass of the pebble, $m = 0.05\ \text{kg}$

Acceleration due to gravity, $g = 9.8\ \text{m/s}^2$ (standard value)

The magnitude of the gravitational force is:

$F_g = mg = (0.05\ \text{kg})(9.8\ \text{m/s}^2) = 0.49\ \text{N}$

The direction of this force is always vertically downwards.

Now, let's address each part of the question:

1. During its upward motion:

   • As the pebble moves upwards, its velocity is directed upwards. However, the only force acting on it is gravity, which is directed downwards.
   • According to Newton's Second Law, the net force determines the acceleration. Since gravity is the only force, the net force is simply the gravitational force.
   • Therefore, the net force is $0.49\ \text{N}$ directed vertically downwards. This downward force causes the pebble to slow down as it moves upwards.

2. During its downward motion:

   • Once the pebble reaches its highest point and starts falling, its velocity is directed downwards.
   • Again, the only force acting on it is gravity, which is directed downwards.
   • The net force is still the gravitational force.
   • Therefore, the net force is $0.49\ \text{N}$ directed vertically downwards. This downward force causes the pebble to speed up as it moves downwards.

3. At the highest point where it is momentarily at rest:

   • This is a common point of confusion. At the highest point, the pebble's vertical velocity momentarily becomes zero.
   • However, being momentarily at rest does not mean the acceleration is zero, nor does it mean the net force is zero. If the net force were zero at the highest point, the pebble would simply hover there, which doesn't happen.
   • The pebble is still under the influence of gravity. Gravity does not "turn off" just because the object stops for an instant.
   • The acceleration due to gravity is still $9.8\ \text{m/s}^2$ downwards.
   • Therefore, the net force (which is $m\vec{a}$) is still the gravitational force.
   • The net force is $0.49\ \text{N}$ directed vertically downwards.
   Watch out

   A common misconception is that the net force at the highest point of projectile motion is zero because the velocity is momentarily zero. Remember that force causes acceleration, not velocity. Even if velocity is zero, if there's an unbalanced force, there will be acceleration.

Do your answers change if the pebble was thrown at an angle of $45^{\circ}$ with the horizontal direction?

No, the answers do not change.

The direction and magnitude of the net force on the pebble remain the same ($0.49\ \text{N}$ vertically downwards) regardless of the angle at which it is thrown. Here's why:

• Gravity's Nature: Gravity is a force that always acts vertically downwards, irrespective of the object's horizontal motion or its trajectory.
• Ignoring Air Resistance: Since we are ignoring air resistance, there are no other forces (like drag) that would depend on the pebble's velocity or direction of motion.
• Components of Motion: When a pebble is thrown at an angle, its motion can be broken down into horizontal and vertical components. The horizontal motion, in the absence of air resistance, has zero acceleration (constant velocity). The vertical motion is solely governed by gravity, causing a constant downward acceleration. The net force is entirely due to this vertical acceleration.

The launch angle only affects the path (trajectory) of the pebble and how long it stays in the air, but it does not change the fundamental force acting on it during its flight.

Concept: Newton's Second Law — gravity is the only unbalanced force once air resistance is ignored

Step 1: Identify every force acting on the pebble after release

With air resistance ignored, the only force acting on the pebble throughout its flight is gravity, $F_g = mg$, always directed vertically downward — this holds during the upward leg, the downward leg, and even at the highest point.

Step 2: Address the highest point specifically

At the highest point the velocity is momentarily zero, but velocity being zero does not mean force is zero — force determines acceleration, and gravity is still acting, so the pebble continues to accelerate downward even as it momentarily pauses.

Step 3: Compute the magnitude

Step 4: Consider the 45° launch case

Splitting the motion into horizontal and vertical components: horizontally there is no force (constant velocity), vertically gravity alone acts exactly as before. The launch angle changes only the trajectory, not the force.

Final Answer:

F = 0.49 N, vertically downward, in all three positions (a),

(b),

(c), and unchanged for a 45° launch.