Unit 5 Mechanical Principles and Applications M2 Work and Energy Part 1

Re-cap on P4 Remember: v=u+at s=ut+1/2at2 v2=u2+2as s=1/2(u+v)t where: s=displacement, u=initial velocity, v=final velocity, a=linear acceleration.

Work When a force F moves a distance x, work done,W, is given by: W=Fx

Types of Energy In the first diagram a mass is raised on a pulley. Work is done and the energy of the mass increases. This energy is stored in the mass as potential energy. In the second diagram a truck is accelerated along a surface. A force is needed to do this and work is done. The energy used to accelerate the truck is stored as kinetic energy.

Potential Energy This is the energy stored in a mass as it is moved upwards against the force of gravity. The weight is mg, the distance moved is h metres (sometimes z or x). W = mgh

Worked example If a mass being lifted is 200 kg and it is raised 0.6 m, determine the work done and the change in P.E. of the mass. Weight = mg, so the force to be overcome = 200 x 9.81 = 1962 N W=F x = 1962 x 0.6 = 1177.2 J

Kinetic Energy This is the energy stored in a mass due to its velocity. F = ma a = v/t x = vt/2 W = fx We can re-arrange all these to get: K.E. = ½ mv2

Worked example A force of 80 N is used to pull a truck 200 m along a horizontal floor. Determine the work done and the increase in K.E. W = Fx = 80 x 200 = 16000 J If no energy is lost due to friction then this must all end up as the Kinetic Energy of the truck.

Conservation of Energy The law of conservation of energy states that energy cannot be created or destroyed, but changed from one form in to another. In many real situations friction causes energy to be “lost” as heat. In the following questions we will assume that there are no losses due to friction.

Falling Bodies A body, which is z metres above a point, has a P.E. of mgz. As it falls P.E. is converted into K.E., ½ mv2. mgz = ½ mv2 2mgz = mv2 v2 = 2gz v = √(2gz)

Worked Example A ball of mass 0.4 kg swings on the end of a tin rod with negligible mass with length 60 mm. The ball is held horizontal and then released. Calculate the following: the velocity of the ball as it passes through its lowest position. the loss of potential energy. the gain in kinetic energy.

Answer i. the ball will swing through a vertical height of 60 mm. v = √(2gz) =√(2 x 9.81 x 0.06) = 1.085 m/s. ii. change in P.E. = mgz = 0.4 x 9.81 x 0.06 = 0.2354 J iii. change in K.E. = ½ mv2 = ½ x 0.4 x 1.0852 = 0.2354 J Note energy is conserved.

Question 1 An object of mass 20 kg is dropped onto a surface from a height of 50 m. Calculate the energy and velocity just before it hits the surface.

Question 2 A swinging hammer must have 50 Joules of energy and a velocity of 2 m/s at the bottom of the swing. Calculate the mass and height of the hammer before it is released.

Question 3 A swinging hammer has a mass of 2 kg and is raised 0.2 m. calculate the energy and velocity at the bottom of the swing.