Question

A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

1. during its upward motion,
2. during its downward motion,
3. at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

Answer 1

When an object is thrown upward or drops downward, the constant gravitational pull of Earth affects it uniformly. This pull results in a constant acceleration a =+g = 10 m/s² downward. If m is the object's mass, then mg is the force acting on it in a downward direction. For a pebble with a mass of m = 0.05 kg, the force is calculated as follows:

1. ∴ Net force on the pebble= mg (∴ a = g)
   = 0.05 × 10 = 0.5 N (acts vertically downwards)
2. ∴ Net force on the pebble = mg (∴ a = g)
   = 0.05 × 10 = 0.5 N (acts vertically downwards)
3. When the stone is at the highest point then also the net force (= mg) acts in vertically downward direction
   ∴ Net force on the pebble = mg
   = 0.05 × 10 = 0.5 N

When the pebble is thrown at a 45-degree angle relative to the horizontal plane, it will develop both vertical and horizontal components of velocity. These components do not influence the gravitational force acting on the pebble. Thus, the calculated force remains unchanged in all scenarios. However, specifically in scenario (c), the pebble is not stationary at its highest point but continues to have a horizontal velocity.