Aim: How do we explain work done by a constant force?

Definition of Work (non-calculus) The work W done by an agent exerting a constant force on a system is the product of the component FcosΘ along the direction of the displacement of the point of application of the force and magnitude of Δx. W=F Δx cos Θ

Characteristics of Work Is work a vector or scalar? Scalar What are the units of work? Joules 3) A person slowly lifts a heavy box of mass m a vertical height h, and then walks horizontally at constant velocity a distance d while holding the box. Determine the work done by (a) by the person and (b) by the gravitational force on the box a) mgh b) -mgh

When will the following occur: Negative work done on an object Positive work done on an object 0 work done on an object The angle between force and displacement must be between 90 and 270 degrees The angle between the force and displacement must either be between 0 and 90 degrees or 270 and 360 degrees 90 degrees and 270 degrees

Thought Question 1 In each of the six situations, a box is being acted upon by forces to the left or to the right. The forces either have magnitudes equal to 1N or 2N. Identify whether the work done by the net force in each case is positive, negative, or zero. Positive Zero Negative

Thought Question 2: Work on Mountain Climber A and Climber B have equal weights. They both climb to the top of the mountain. Climber A takes a path which has a gentle slope. Climber B takes a path which has a steep slope. Ignoring friction, who does more work in climbing the mountain? SAME WORK

Mr. Clean problem A man cleaning his apartment pulls a vacuum cleaner with a force of magnitude F = 50.0 N. The force makes an angle of 30 degrees with the horizontal. The vacuum cleaner is displaced 3.00 m to the right. Calculate the work done by the 50 N force on the vacuum cleaner.

Work Problems 2) Find the work done by the man on the vacuum cleaner if he pulls 3.00 m with horizontal force of 32.0N. 3) If a person lifts a 20 kg bucket slowly from a well and does 6000J work work on the bucket, how deep is the well? 4) A 65 kg woman climbs a flight of 20 stairs, each 23 cm high. How much work is done by the gravitational force in the process? 2) 96 J 3) 30.6 m 4) -2.93 kJ

Work-Energy Theorem When work is done on a system and the only change in the system is in its speed, the work done by the net force equals the change in kinetic energy of the system. W = ΔKE where KE = ½ mv2

Thought Question 3 Rank the following velocities according to the kinetic energy the particle will have, greatest first: a) v=4i + 3j b) v=-4i+3j c) v=-3i+4j d) v=3i-4j e) v=5i f) v= 5 m/s at 30 degrees to the horizontal Each particle has the same kinetic energy

Work-Energy Thm Example A 6 kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant, horizontal force of 12.0 N. Find the speed of the block after it has moved 3 m. A 0.6 kg particle has a speed of 2.00 m/s at point A and kinetic energy of 7.50 J at point B. What is its kinetic energy at A? It speed at B? The total work done on the particle as it moves from A to B? 3.46 m/s

Dot Product of Two Vectors W = F ˚Δr Work is the dot product of a force vector and displacement vector. W=|F||r|cosᶿ or W=Fxrx +Fyry What is true about the dot product of two vectors and the product of the magnitudes of the two vectors?

Dot product Examples 1. A force F = 2i + 3j causes an object to have a displacement, Δr = -i + 2j. Calculate the work done on the object. Find the angle Θ between the force and displacement vectors. 60.5 degrees

Dot Product Examples 2) As a particle moves from the origin to (3i – 4j) m, it is acted upon by a force given by (4i-5j) N. Calculate the work done by this force. W=Fxrx +Fyry = (3)(4) + (-4)(-5) = 32 J 32 J

Thought Question 4 Is positive or negative work done by a constant force F on a particle during a straight line displacement d if (a) the angle between F and d is 30 degrees;(b) the angle is 100 degrees;(c)F=2i-3j and d=-4i Positive Work Negative Work

Conclusion A 3 kg object has an initial velocity vi =(6i-2j) m/s What is its kinetic energy at this time? Find the total work done on the object if its velocity changes to (8i+4j)m/s v= √40 so KE=1/2mv2=1/2(3)(40)= 60 J V=√80 so KE=1/2mv2=1/2(3)(80) =120 J So 60 J of work must have been done because this was the change in kinetic energy a) 60 J b) 60 J