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

For all the scenarios listed (a to e), the net force acting on the object is zero. This is because each object is either at rest or moving with a constant velocity, which implies zero acceleration. According to Newton's second law of motion, the net force on an object is directly proportional to its acceleration ($F\; =\; m\; \backslash times\; a$). If the acceleration is zero, the net force must also be zero.

a) A rain drop falling with constant speed has zero acceleration, hence the net force is zero. The downward force of gravity is balanced by the upward force of air resistance.

b) A cork floating on water is in equilibrium. Its weight acting downwards is balanced by the upward buoyant force exerted by the water. Thus, the net force is zero.

c) A kite held stationary in the sky is at rest, meaning its acceleration is zero. Therefore, the net force acting on it is zero. The forces like lift, drag, and tension are balanced.

d) A car moving with a constant velocity has zero acceleration. According to Newton's second law, the net force acting on the car is zero. The forward driving force is balanced by the opposing forces like friction and air resistance.

e) A high-speed electron in space, far from all fields and forces, experiences no external influence. Thus, the net force acting on it is zero, and it continues to move with its constant velocity.

The net force acting on the pebble in all cases (upward motion, downward motion, and at the highest point) is the gravitational force, which acts vertically downwards. Air resistance is ignored as per the question.

Given:

• Mass of the pebble, $m\; =\; 0.05\; \backslash text\{\; kg\}$
• Acceleration due to gravity, $g\; =\; 10\; \backslash text\{\; m/s\}^2$ (approximately, acting downwards)

Using Newton's second law of motion, the net force ($F$) is calculated as:

$$F\; =\; m\; \backslash times\; a$$

In this case, the acceleration ($a$) is the acceleration due to gravity ($g$), which is always directed downwards.

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

$$F\; =\; 0.5\; \backslash text\{\; N\}$$

Therefore, the magnitude of the net force is 0.5 N, and its direction is vertically downwards.

a) During its upward motion, the net force is 0.5 N downwards.

b) During its downward motion, the net force is 0.5 N downwards.

c) At the highest point, where the pebble is momentarily at rest (vertical velocity is zero), the net force is still 0.5 N downwards due to gravity.

Change if thrown at an angle:

If the pebble is thrown at an angle of 45° with the horizontal, it undergoes projectile motion. The net force acting on it at any point during its flight (ignoring air resistance) is still the gravitational force, which is 0.5 N acting vertically downwards. The horizontal component of velocity does not affect the net vertical force, which is solely due to gravity.