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Published byEvelyn Harris Modified over 9 years ago
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Work and energy
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Objectives 1.Recognize the difference between the scientific and the ordinary definitions of work. 2.Define work, relating it to force and displacement. 3.Identify where work is being performed in a variety of situations. 4.Calculate the net work done when many forces are applied to an object. Homework:
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In physics, work is defined as a force acting upon an object to cause a displacement. Definition and Mathematics of Work Work is being done Work is not being done
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Let’s practice – work or no work 1.A student applies a force to a wall and becomes exhausted. 2.A calculator falls off a table and free falls to the ground. 3.A waiter carries a tray full of beverages above his head by one arm across the room 4.A rocket accelerates through space. no work work no work work
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Calculating the Amount of Work Done by Forces θ F d F - is the force in Newton, which causes the displacement of the object. d - is the displacement in meters θ = angle between force and displacement W - is work in N∙m or Joule (J). 1 J = 1 N∙m = 1 kg∙m 2 /s 2 Work is a scalar quantity Work is independent of time the force acts on the object. θ F d FxFx FyFy Only the horizontal component of the force (Fcosθ) causes a horizontal displacement.
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Example 1 How much work is done on a vacuum cleaner pulled 3.0 m by a force of 50.0 N at an angle of 30 o above the horizontal?.
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Example 2 How much work is done in lifting a 5.0 kg box from the floor to a height of 1.2 m above the floor? W = F∙dcosθ F = mg = (5.0 kg)(9.81 m/s 2 ) cos0 o = 49 N W = F∙d = (49 N) (1.2 m) = 59 J Given: d = h = 1.2 meters; m = 5.0 kg; θ = 0 Unknown: W = ?
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Example 3 A 2.3 kg block rests on a horizontal surface. A constant force of 5.0 N is applied to the block at an angle of 30. o to the horizontal; determine the work done on the block a distance of 2.0 meters along the surface. Given: F = 5.0 N; Given: F = 5.0 N; m = 2.3 kg d = 2.0 m θ= 30 o 30 o 5.0 N 2.3 kg unknown: unknown: W = ? J W = ? J Solve: Solve: W = F∙d∙cos W = F∙d∙cosθ W = (5.0 N)(2.0 m)(cos30 o ) = 8.7 J W = (5.0 N)(2.0 m)(cos30 o ) = 8.7 J
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Example 4 Matt pulls block along a horizontal surface at constant velocity. The diagram show the components of the force exerted on the block by Matt. Determine how much work is done against friction. 8.0 N 6.0 N 3.0 m W = F x d x W = (8.0 N)(3.0 m) = 24 J Given: F x = 8.0 N Given: F x = 8.0 N F y = 6.0 N d x = 3.0 m unknown: W = ? J unknown: W = ? J F
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Class work Page 170 practice #1-4 1.1.50 x 10 7 J 2.7.0 x 10 2 J 3.1.6 x 10 3 J 4.1.1 m
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The sign of work
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When No work is done
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Force vs. displacement graph The area under a force versus displacement graph is the work done by the force. Displacement (m) Force (N) work Example: a block is pulled along a table with 10. N over a distance of 1.0 m. W = Fd = (10. N)(1.0 m) = 10. J height basearea
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The angle in work equation The angle in the equation is the angle between the force and the displacement vectors. F & d are in the same direction, θ is 0 o. F d
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What is θ in each case?
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Class work Page 171 #1-6
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