CBSE Class 11 Physics Chapter 5: Laws of Motion NCERT Solutions

The net force acting on an object is determined by its acceleration, as per Newton's second law of motion ($F\; =\; ma$). If an object is moving with constant speed or constant velocity, its acceleration is zero, which implies the net force acting on it is also zero.

1. a) A drop of rain falling with constant speed: Since the speed is constant, the acceleration is zero. Therefore, the net force acting on the rain drop is zero.
2. b) A cork of mass 10 g floating on water: For the cork to float, the upward buoyant force exerted by the water must be exactly balancing the downward force of gravity (its weight). This means the net force acting on the cork is zero.
3. c) A kite held stationary in the sky: If the kite is stationary, it means it is not moving at all, so its velocity and acceleration are both zero. According to Newton's first law, an object remains at rest unless acted upon by a net external force. Thus, the net force acting on the kite is zero.
4. d) A car moving with a constant velocity of 30 km/h on a rough road: A constant velocity implies that the acceleration of the car is zero. Therefore, as per Newton's second law, the net force acting on the car is zero.
5. e) A high-speed electron in space far from all material objects, and free of electric and magnetic fields: If the electron is far from all objects and free of any fields, there are no forces acting on it. Hence, the net force acting on the electron is zero.

We are given the mass of the pebble, $m\; =\; 0.05\; \backslash text\{\; kg\}$. We need to find the net force acting on the pebble in different stages of its motion, ignoring air resistance. The only force acting on the pebble in the absence of air resistance is the force of gravity.

The force of gravity (weight) is given by $W\; =\; m\; \backslash times\; g$, where $g$ is the acceleration due to gravity, approximately $10\; \backslash text\{\; m/s\}^2$. This force always acts vertically downwards.

The magnitude of the gravitational force is:

$$F\; =\; m\; \backslash times\; g\; =\; 0.05\; \backslash text\{\; kg\}\; \backslash times\; 10\; \backslash text\{\; m/s\}^2\; =\; 0.5\; \backslash text\{\; N\}$$

The direction of this force is always vertically downwards.

a) During its upward motion: Even though the pebble is moving upwards, the only force acting on it is gravity, which acts downwards. Therefore, the net force on the pebble is 0.5 N, acting vertically downwards.

b) During its downward motion: As the pebble falls downwards, the force of gravity continues to act on it in the downward direction. Thus, the net force on the pebble is 0.5 N, acting vertically downwards.

c) At the highest point where it is momentarily at rest: At the highest point, the vertical velocity of the pebble is momentarily zero. However, the force of gravity is still acting on it. Therefore, the net force on the pebble at this point is 0.5 N, acting vertically downwards.

Change in answers if thrown at an angle of 45°:

If the pebble is thrown at an angle of 45° with the horizontal, it becomes a projectile. The motion can be resolved into horizontal and vertical components. The force of gravity acts only in the vertical direction. Air resistance is ignored. Therefore, the net force acting on the pebble throughout its trajectory (upward, downward, or at the highest point) remains the gravitational force, which is 0.5 N acting vertically downwards. The horizontal component of velocity does not affect the net vertical force.

Conclusion: The direction and magnitude of the net force on the pebble remain the same (0.5 N, vertically downwards) in all cases, regardless of its direction of motion or whether it was thrown vertically or at an angle, as long as air resistance is ignored.