Unit Three: Energy Chapter 4 and 6 Work p. 181 1-7 extra p. 183 1-7 p. 226 1-4, 9-11 Booklet p.23 1-4 p.24 13, 14, 18, 19, 24, 26, 30 Kinetic Energy and.

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Unit Three: Energy Chapter 4 and 6 Work p extra p p , 9-11 Booklet p p.24 13, 14, 18, 19, 24, 26, 30 Kinetic Energy and the Work-Energy Theorem p ,2,4-8extrap p ,6,12,13 Booklet p , 20-23, 25, 27-29

Work The work done by a constant force is defined as the product of the component of the force in the direction of the displacement and the magnitude of the displacement.

W = Work units of J or Nm or the Joule is named after James Prescott Joule  is the angle between the force and the displacement

Note that Fcos  is the component of the force in the direction of the displacement. If the angle is greater than ninety degrees then the work will be negative (cos .

Work is a scalar quantity. Energy is defined as the ability to do work and therefore is a scalar quantity as well. Work can be positive or negative but these signs are not direction. We will see that they indicate a gain of kinetic energy or a loss of kinetic energy respectively.

Negative work is done on an object when it is slowed by a force. Positive work is done when an object is sped up by a force. The area under a F-d graph is equal to the work done by an applied force. Assume the force and displacement are colinear.

The total work done on an object is the sum of all the work done by individual forces.

Rousseau pushes with a force of 500 N on an immovable wall. How much work is done on the wall? David swings a rock around his head with a centripetal force of 250 N. The rock goes around his head 3 times in 0.56 s (the radius of the circle is 0.8 m). What is the work done on the rock?

A 4 kg block is raised 5 m. How much work is done on the block? assume constant v

This means the work done on the ball is 0 J. note

A sled (15 kg) is pulled with a 50 N [20 o ath] force for 7.5 m. The coefficient of friction is Calculate the work done by each force and total work done on the sled. 20 o

Work done by F N and F g are zero since they are perpendicular to the displacement. Work done by F a

To calculate work done by friction we must calculate F N.

Work done by F f Therefore the total work done on the sled is J

WORK ENERGY THEOREM For an object that is accelerated by a constant net force and moves in the same direction...

define kinetic energy as (E k) The work done by the net force acting on a body is equal to the change in the kinetic energy of the body.

A ball is dropped from rest at a height of 8.25 m. What will be its speed when it hits the ground? (could solve this kinematically but let’s do it using the work-energy theorem. divide by m

Therefore the speed of the ball is m/s when it hits the ground.

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