RI Y5 Remedial Dynamics.1 Assignment

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 1 Chapter 3: Dynamics (Newton’s Laws of Motion) Newton’s First Law of motion: A body continues in its state of rest or uniform motion in a straight line, unless a resultant external force acts on it. Newton’s Second Law of motion: The rate of change of momentum of a body is proportional to the resultant force acting on it and occurs in the direction of the force. Newton’s Third Law of motion: If a body A exerts a force on body B, then body B exerts an equal and opposite force on body A. Worked Example 1 (Newton’s Second Law) A light string connects two blocks X and Y, of mass 5.0 kg and 20 kg respectively. A force of 50 N pulls them along a smooth horizontal surface. Determine the force exerted on X by the string. Draw FBD for each object: Both X and Y experience the same acceleration a. Consider the system as a whole (X and Y together), F = (m x + my) a 50 = (5 + 20) a a = 2.0 m s–2 Next consider only block X, T = m x a = (5)(2.0) = 10 N Hence the force on X by the string is 10 N. X 5.0 kg Y 20 kg 50 N 50 N a T T X 5.0 kg a Y 20 kg

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 2 Worked Example 2 An object of mass 800 kg is sliding down a slope inclined at an angle of 30  to the horizontal. Sketch the forces acting on the object and calculate the acceleration of the object if: (a) the slope is frictionless (b) the frictional force acting on the object is 1000 N. (a) Object on slope without friction Taking downward along the slope as positive. Consider forces acting along the slope: Resolving the weight of the object in the direction along the slope (i.e. resultant force on the object along slope): W// = W sin  Using Newton’s Second Law: F = ma (i.e. resultant force on the object along the slope will cause the object to accelerate) F = ma mg sin  = ma a = g sin  = 9.81 sin 30 = 4.9 m s2 (b) Object on slope with friction Taking downward along the slope as positive. Consider forces acting along the slope: Using Newton’s Second Law: F = ma mg sin  – f = ma (800)(9.81) sin 30 – 1000 = (800) a a = 3.7 m s2  W N f  W N

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 3 Worked Example 3 A man of mass 65 kg stands on a weighi ng scale in an elevator going to the 25 th storey. The elevator accelerates from the ground floor at 3.0 m s2 from rest to speed v. Once it reaches speed v, it continues moving at constant speed v. When approaching its destination, it decelerates at 2.0 m s 2 till rest. Determine the reading, i