A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,(a) during its upward motion,(b) during its downward motion,(c) 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. from Physics Laws of Motion Class 11 Manipur Board
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A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) 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.


Irrespective of the direction of motion, acceleration due to gravity always acts downward.

The gravitational force is the only force that acts on the pebble in all three cases.

Its magnitude is given by Newton’s second law of motion as:

                       F = m x a

where,
 
F = net force, 

m= mass of the pebble, and 

a is the acceleration.

Here, a = g = 9.8 m/s2

Therefore, 

               F = 0.05 x 10 = 0.5 N.

The magnitude of net force = 0.5 N, which is acting vertically in the downward direction.

If the pebble is thrown at an angle of 45° with the horizontal direction, it will have both the horizontal and vertical components of velocity. At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble. 
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