11141_n_24745(6)

# 11141_n_24745(6) - 1 ENERGY Energy present in a variety of...

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Unformatted text preview: 1 ENERGY Energy present in a variety of forms Mechanical energy Chemical energy Nuclear energy Electromagnetic energy Energy can be transformed form one form to another Energy is conserved (isolated system) WORK Energy is the ability to do WORK Work done by a constant force F Direction of Motion θ Displacement = ∆ x F F F Work = Force x displacement W = (Fcos θ ) ∆ x θ = angle between the force and the displacement Case 1: θ = 0 W = (Fcos0 ) ∆ x = F ∆ x > 0 Case 2: θ = 180 W = (Fcos180 ) ∆ x = - F ∆ x <¡¡0 Case 3 (general): θ W = (Fcos θ ) ∆ x = F // ∆ x F / = component of F parallel to displacement (Friction) F // WORK Work = Force x displacement Scalar SI Units: SI Units: (Newton)(meter) = Nm Nm or Joules ( J ) What is the work done by done by F? • θ = 90 so W = (Fcos90 ) ∆ x = 0 J Displacement = ∆ x F Direction of Motion 2 Work Done by a Constant Force Displacement d F θ 0 < θ < 90 0 W > 0 90 0 < θ < 180 W < 0 270-1 180 90 1 cos θ θ Examples Examples Problem 8: Problem 8: A block of mass 2.50 kg is pushed 2.20 m along a frictionless frictionless horizontal table by a constant 16.0 N force directed 25 below the horizontal. Determine a) the work done by the applied force: W F = (16.0 cos25 N)(2.20 m) = 31.9 Nm = 31.9 J b) the work done by the normal force: W N = (N cos90 N)(2.20 m) = 0 J c) the work done by the force of gravity W g = (mg cos90 N)(2.20 m) = 0 J d) The work done by the net force on the block W net = (16.0 cos25 N)(2.20 m) = 31.9 J or W net = W F + W N + W g = 31.9 + 0 + 0 = 31.9 J F θ ∆ x = 2.20 m W N Examples Examples Problem 5: Problem 5: Starting from rest, a 5.00 kg block slides 2.50 m down a rough 30Starting from rest, a 5....
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11141_n_24745(6) - 1 ENERGY Energy present in a variety of...

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