(a)F = ma, rearranged to a = F/m.

(b)Newton's Third Law: when object A exerts a force on object B, object B exerts an equal and opposite force on object A.

(a)F = kx, with x converted to metres (6.0 cm = 0.06 m).

(b)Hooke's Law: extension is directly proportional to the applied force, provided the elastic limit is not exceeded.

(a)Weight (400 N) acting downward, and the normal contact force from the floor acting upward.

(b)The box is in equilibrium (not accelerating), so by Newton's First Law the resultant force must be zero — the normal force must balance the weight exactly.

(a)Initially weight exceeds air resistance, so the diver accelerates downward. As speed increases, air resistance increases. Once air resistance equals weight, the resultant force is zero, so acceleration is zero and velocity stays constant — terminal velocity.

(a)a = F/m.

(a)k = F/x, with x converted to metres.

(a)For perpendicular forces, use Pythagoras: resultant = √(F₁² + F₂²).

(a)The component of weight down a slope is W sin θ.

(a)Rearranging F = ma gives m = F/a.

(a)An object is in equilibrium when the resultant of all forces acting on it is zero, so it is either at rest or moving at constant velocity.

(a)The table exerts a normal contact force upward that exactly balances the cup's weight downward, so by Newton's First Law the resultant force is zero and the cup stays at rest.