Ans: One newton is the net force that gives a 1 kg object an acceleration of 1 m/s². Thus, 1 N = 1 kg m/s².

Ans: Balanced forces have zero net force and do not change velocity: the object remains at rest or continues with constant velocity. Unbalanced forces have a non-zero net force and produce acceleration, changing speed, direction, or both.

Ans: Friction from the ground acts forwards on the supporting foot. The foot pushes backwards against the ground and tends to slip backwards, so friction opposes that tendency and helps propel the child forwards.

Ans: Padding compresses and increases the time over which the person stops. For the same change in velocity, a longer stopping time reduces the magnitude of average acceleration, since a = (v − u)/t. For the same mass, F_net = ma then gives a smaller average stopping force, reducing the risk of injury.

Ans: The horizontal net force is 18 − 8 = 10 N forwards. Acceleration a = F_net/m = 10/5 = 2 m/s² forwards. Using s = ut + ½at² with u = 0, displacement s = ½ × 2 × 3² = 9 m forwards.

Ans: Choose both boxes and the string as one system. The string tension is internal: it pulls the two boxes in opposite directions, and these contributions cancel in the system’s total force. The external horizontal force is 18 N, while weight and support forces balance vertically. Total mass = 2 + 4 = 6 kg, so a = 18/6 = 3 m/s² in the pulling direction. For the 4 kg box alone, tension is its only horizontal force. Therefore, T = ma = 4 × 3 = 12 N.