Newton’s First Law

1. Newton’s First Law Balanced Forces mean Constant Speed in a Straight Line Steady Speed in a Straight Line means Balanced Forces Here is a man being winched to safety by a helicopter rescue line. His mass is 70 kg and he is being raised at a speed of 3 m/s. What is the tension force in the line? T There are two forces acting on the man, which must be equal if he is moving up at a steady speed. W T = W W = mg = 70 x 10 = 700 N Tension in rescue line = 700 N Note: the tension would be the same if the man was being lowered at a steady speed or if he was being held stationary.

2. Newton 2- part 1 • Acceleration increases with increase in unbalanced force • Acceleration decreases with increase in mass Both vehicles have the same engine, but different masses give different acceleration.

3. Newton 2 - part 2 The mass of this cyclist and his bike is 70 kg. If he is to accelerate at 1.5 m/s2, what unbalanced force must act on him and his bike? F = ma m= 70 kg = 70 x 1.5 a= 1.5 m/s2 = 105 N

4. Newton 2 - part 3 What forces are acting on the cyclist (and his bike) ? Pedalling force Friction forces If pedalling force = friction forces cyclist goes at steady speed cyclist accelerates If pedalling force > friction forces cyclist decelerates If pedalling force < friction forces

5. Newton 2 - part 4 Acceleration in m/s2 Unbalanced force in Newtons F = ma Mass in kg Fr E Unbalanced force which accelerates car forward = engine force E - friction Fr E Unbalanced force which accelerates rocket upwards = engine thrust E - weight W W

6. speed D C B time A Newton 1 and 2 AR W A to B Constant acceleration : constant force : AR = 0 B to C Decreasing acceleration : unbalanced force decreasing i.e. Air resistance increasing. Constant speed : balanced forces : weight = air resistance C to D

7. Work 1 Work is done when a force moves through a distance and energy is transformed. The work done (and the energy transformed) is equal to the force multiplied by the distance moved in the force’s direction. distance moved by force in metres Ew = F x d work done in joules force acting in Newtons

8. Work 2 Bruiser Fido Fido and Bruiser are both exerting a force. They are not doing any work, however, because they are not moving through any distance. Remember:- work equals force x distance.

9. Work 3 25 N 2 m Work done = F x d = 25 x 2 = 50 J

10. 2 m 4 kg Work 4 Doing work against gravity means using a force to balance the downwards force of gravity, namely, weight. For this man to lift the 4 kg box, he must use a force equal to the weight of 4 kg. Since W = mg, he needs 4 x 10 N Work done = force x distance = 40 x 2 = 80 J

11. Work 5 - Lifting Things When an object is lifted, a force equal to the object’s weight must be used. Example: A boy drags a 5 kg box along the floor for 2 m, using a 20 N force. Then he lifts it on to a ledge 1 m above the floor. How much work does he do? Work done dragging = force x distance = 20 x 5 = 100 J Work done lifting = force x distance = weight x height weight of 5 kg = mg = 5 x 10 = 50 N so work done lifting = 50 x 1 = 50 J Total work done = 100 + 50 = 150 J

12. h metres Work 6- Potential Energy total mass raised = m kg force needed = mg Newtons work done = F x d = mg x h Work done is stored as potential energy = mgh joules