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Force: is a push or a pull. There are contact forces such as pushing, pulling, or friction. There are field forces such as gravitation and electrostatics. Force is measured in Newtons, N.

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Force, Mass, and Acceleration Isaac Newtons 2 nd law of motion states: the acceleration of an object is directly proportional to the net force on an object and indirectly proportional to the acceleration of an object. That is, a = F net / m The net force, F net, is defined as the sum of all forces acting along a given line of action. Example: 7N+ -4N = 3N In this example, 3N is the net force.

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Weight: is the attractive force between any mass and the mass of the earth. Weight = mass x gravitational acceleration F W = m g where g = -10 m/s/s So, the weight of 1 kilogram is -10 Newtons (N). F W = m g F W = (1 kg) (-10 m/s/s) = -10 N

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Find: A) the weight of the block. B) the force applied to the block, if the acceleration of the block is 0 m/s/s. C) the net force on the block. D) the force applied to the block, if the acceleration of the block is 1.2 m/s/s. to the right. F = ? Friction = 5N M = 10 kg

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Find: A) the weight of the block. (W = 98 N) B) the force applied to the block, if the acceleration of the block is 0 m/s/s. (F = 5N) C) the net force on the block. (Fnet = 0 N) D) the force applied to the block, if the acceleration of the block is 1.2 m/s/s. to the right. (F = 17 N) F = ? Friction = 5N M = 10 kg

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Find A) the weight of the block. B) the tension in the rope tied to the block, if the acceleration of the block is 1.2 m/s/s upwards. T = Weight = M = 10 kg

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Find A) the weight of the block. (W = 98 N) B) the tension in the rope tied to the block, if the acceleration of the block is 1.2 m/s/s upwards. ( T = 110 N) T = Weight = M = 10 kg

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Newtons Laws Practice Probs. (accelerating objects) Newtons Laws Practice Probs. (accelerating objects) 1. Calculate the force that must be applied to produce an acceleration of 1.8 gs (18 m/s/s) on a 1.2 kg puck sliding free of friction on an air table.

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Newtons Laws Practice Probs. Newtons Laws Practice Probs. 1. Calculate the force that must be applied to produce an acceleration of 1.8 gs (18 m/s/s) on a 1.2 kg puck sliding free of friction on an air table. F = 21.6 N

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2 What will be the acceleration of a sky diver when air resistance is half the weight of the sky diver?

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2 What will be the acceleration of a sky diver when air resistance is half the weight of the sky diver? 0.5g

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Find: A)The weight of the 1200 kg elevator. B)The upward force on the elevator when it is pulled upward at 2 m/s/s.

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Find: A)The weight of the 1200 kg elevator. F = W = (1200 kg)(-10 m/s/s) = -12,000 N B)The upward force on the elevator when it is pulled upward at 2 m/s/s. Fnet = m a F - 12,000 N = (1200 kg)(2 m/s/s) F = 2,400 N +12, 000 N F = 14, 400 N

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A 2 kg hanging block accelerates the 8 kg block on the frictionless table. What is the acceleration of the two block system?

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A 2 kg hanging block accelerates the 8 kg block on the frictionless table. A.What is the acceleration of the two block system? a = 20 N/10 kg = 2 m/s/s

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A 2 kg hanging block accelerates the 8 kg block on the frictionless table. What is the tension in the string supporting the 2 kg mass?

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A 2 kg hanging block accelerates the 8 kg block on the frictionless table. The tension, T, in the string on the 2kg mass F net = ma T - (2 kg)(10 m/s/s) = (2 kg)(-2 m/s/s) T = 20 N - 4 N T = 16 N

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Find the rate of acceleration of the 8 kg and 2 kg blocks. What is the tension in the string?

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Mg – T = Ma T – mg = ma Adding these yields: Mg – mg = (M + m) a a = g(M –m) / (M +m) so a =- 10 (6/10) = -6 m/s/s T T mg Mg The acceleration of mg is down (negative), while the acceleration of Mg is up (negative) as both are connected via the string.

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To find the tension in the string ( which is the same throughout the string), we may use: T – mg = ma T = ma + mg T = m(a + g) = 2(-6 + 10) = 8 N Note: g is NOT -10 as the sign is part of the eqtn. T T mg Mg The acceleration of mg is down (negative), while the acceleration of Mg is up (negative) as both are connected via the string.

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Chapters 5, 6 Force and Laws of Motion. Newtonian mechanics Describes motion and interaction of objects Applicable for speeds much slower than the speed.

Chapters 5, 6 Force and Laws of Motion. Newtonian mechanics Describes motion and interaction of objects Applicable for speeds much slower than the speed.

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