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Newton’s Second Law

Newton’s Second Law. 12.2. Newton’s first law of motion describes the motion of an object with no net force acting on it. Net Force is the sum of all the forces. If all the forces balance each other the net force is zero. Newton’s second law of motion

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Newton’s Second Law

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  1. Newton’s Second Law 12.2

  2. Newton’s first law of motion describes the motion of an object with no net force acting on it.

  3. Net Force is the sum of all the forces. • If all the forces balance each other the net force is zero

  4. Newton’s second law of motion • Newton’s second law of motion states “An object acted upon by a net force will accelerate in the direction of the force”. The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

  5. This verbal statement can be expressed in equation form as follows: a = Fnet / m • The above equation is often rearranged to a more familiar form as shown below. The net force is equated to the product of the mass times the acceleration. Fnet = m * a • a unit of force is equal to a unit of mass times a unit of acceleration.

  6. Examples: • 1. Determine the accelerations which result when a 12 N net force is applied to a 3 kg object and then to a 6 kg object. a = Fnet / m a= 12/3=4 m/s2 a=12/6=2 m/S2

  7. 2. A net force of 15 N is exerted on an encyclopedia to cause it to accelerate at a rate of 5 m/s2. Determine the mass of the encyclopedia. • Fnet= m * a with Fnet = 15 N and a = 5 m/s2 (15 N) = (m)*(5 m/s2) • And m = 3.0 kg 

  8. 3. Suppose that a sled is accelerating at a rate of 2 m/s2. If the net force is tripled and the mass is doubled, then what is the new acceleration of the sled? • 3 m/s2 • The original value of 2 m/s/s must be multiplied by 3 (since a and F are directly proportional) and divided by 2 (since a and m are inversely proportional)

  9. 4. Suppose that a sled is accelerating at a rate of 2 m/s2. If the net force is tripled and the mass is halved, then what is the new acceleration of the sled? • 12 m/s2 • The original value of 2 m/s/s must be multiplied by 3 (since a and F are directly proportional) and divided by 1/2 (since a and m are inversely proportional)

  10. 5-An applied force of 50 N is used to accelerate an object to the right across a frictional surface. The object encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the mass, and the acceleration of the object. (Neglect air resistance.)

  11. 80 N 8 Kg 4.9 m/s2 40 N, right

  12. Since there is no vertical acceleration, normal force = gravity force. • The mass can be found using the equation Fgrav = m • g. m= 80 /10= 8 Kg • The Fnet is the vector sum of all the forces: For vertical forces: 80 N, up plus 80 N, down equals 0 N. • For horizontal forces : 50 N, right plus 10 N,left = 40 N, right.

  13. Finally, a = Fnet / m = (40 N) / (8.16 kg) = 4.9 m/s2

  14. Newton’s Second Law and Momentum • What happens in terms of motion when a net force acts on an object? Its momentum changes. • The change in momentum P2- P1 is the difference between the object’s momentum before and after the force acts. This is described by the following formula:  Force * time = Change in momentum

  15. We know • P1= mv1 ; P2= mv2 substitute in the above equation and you get: • Ft= mv2- mv1 • So the force that acts on an object is related to the initial velocity, and final velocity of the object.

  16. Example 1: A ball of mass 5 kg is moving at 2m/sec. A force takes 4 seconds to stop the ball. Find the force. • Ft= mv2- mv1 • F *4 = (5)*(0)-(5)*(2) • F= -2.5 N

  17. Example2: A hockey stick exerts a force of 100 N on a 0.2 kg puck for 0.1 second. What is the final speed of the puck? (refer to page 338 figure 12-13) • Ft= mv2- mv1 • (100)*( 0.1)= (0.2 )v2 – (0.2)* (0) • v2 = 50 m/s

  18. Example 3: When you catch a ball you exert a force opposite to the direction of the ball’s motion. If you exert a force of 40 N on a 0.5 kg ball moving at 10 m/s, how long does it take to stop the ball?(Book pg 339 figure 12-14) • Ft= mv2- mv1 • (-40) * t = (0.5)*(0)-(0.5)*(10) • t=0.125 seconds

  19. Example 4: A bicyclist experts a force of 24 N and accelerated from 2 m/s to 4 m/s. • a- Find the time needed to do this if the bicycle has a mass of 90 kg. Ft= mv2- mv1 (24) * t = (90)*(4)-(90)*(2) t= 7.5 seconds

  20. b- The same bicycle exerts a force on the brakes to slow from 4 m/s to 0 m/s. this takes 12 seconds. What was the force? In the force positive or negative? • Ft= mv2- mv1 • F(12)= (90)( 0-4) • F(12)= -360 • F= - 30 N

  21. Warm up 1-As long as the forces on an object are balanced, the object will tend to continue moving at what velocity? constant 2- What is the force that brings most objects to rest? Friction 3- When a force is applied to on object for a period of time, the object experiences a change in -------- Velocity or momentum 4- When an objects is sitting at rest on a table, the force of gravity is balanced by --------------- Normal force

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