Work and Energy

Work Physics definition of Work: Work : is the product of the magnitudes of the component of force along the direction of displacement and the displacement W = Fd F=ma W = m a d W = work F = force D = displacement

Work Work is done only when components of a force are parallel to a displacement. displacement force WORK! force displacement NO WORK!

Is Work Happening? A tug of war that is evenly matched A student carries a bucket of water along a horizontal path while walking at a constant velocity. A Crane lifting a car. A person holding a heavy chair at arm’s length for several minutes. A train engine pulling a loaded boxcar initially at rest.

Work Units Work = F x d Work = m x a x d Work = (newtons) (m) (Newton x m) = joules (J)

Work Problem 1 A tugboat pulls a ship with a constant horizontal net force of 5.00 x 10 3 N. How much work is done if the ship is pulled a distance of 3.00 km? W = F x d = 5.00 x 10 3 N ( 3000 m) = 1.5 x 10 7 Nm or J

Work Problem 2 If 2.0 J of work is done raising a 180 g apple, how far is it lifted? W =Fd F = mg = 0.18kg(9.8 m/s 2 ) = 1.76 N W = Fd 2 J = (1.76N)d d = 1.1 m

Work Problem 3 A weight lifter lifts a set of weights a vertical distance of 2.00 m. If a constant net force of 350 N is exerted on the weights, what is the net work done on the weights? W = Fd W = 350 N x 2.00 m = 700 Nm or 700J

Sample Problem 4 What work is done by a forklift raising 583 kgs of frozen turkeys 1.2 m? W = Fd F = 583 kg (9.8 m/s 2 ) = N W = N ( 1.2m) = 6856 Nm or 6856 J

Problems with Forces at Angles θ F Fx = Fcos θ Fy = Fsin θ Because the displacement of the box is only in the x direction only the x-component of the force does work on the box. W = Fdcos θ X-direction

Sample Problem A sailor pulls a boat a distance of 30.0 m along a dock using a rope that makes an angle of 25 o with the horizontal. How much work is done if he exerts a force of 255 N on the rope? 255 N 25 0 W = Fdcos θ = 255N(30m)cos25 = 6.93 x 10 3 J

Sliding up an Incline What we calculated was... For sliding an object up an incline.. W = Fd W = (mg sinθ) d

Sample Problem An airline passenger carries a 215 N suitcase up stairs, a displacement of 4.20 m vertically and 4.60 m horizontally. How much work does the passenger do? 4.6 m 4.2 m

Sample Problem 4.6 m 4.2 m First have to calculate hypotenuse and θ Tan θ = 4.2 m/ 4.6 m Θ = Hypotenuse 2 = A 2 + B 2 Hypotenuse 2 = (4.2) 2 + (4.6) 2 Hypotenuse = 6.23 m m

Sample Problem 4.6 m 4.2 m 6.23 m F║ = mg sin θ = 215 sin 42.4 = 145 N W = Fd = 145 N ( 6.23 m) = 899 Nm = 903 J This should equal the force of the suitcase moving it vertically 4.2 m W = 215 N ( 4.2 m) = 903 J Suitcase weighs 215 N

Graphs of Force vs. Displacement Displacement Force Displacement Force Work = Fd Work can be found graphically by finding the area under the curve

Homework Do Work/Energy/Power worksheet #1-4