Sect. 7.5: Kinetic Energy Work-Kinetic Energy Theorem

Energy  The ability to do work Kinetic Energy  The energy of motion “Kinetic”  Greek word for motion An object in motion has the ability to do work

Object undergoes displacement Δr = Δx i (Δx= x f - x i ) & velocity change (Δv= v f -v i ) under action of const. net force ∑F figure  Text derivation Calculus not needed! Instead, Newton’s 2 nd Law ∑F = ma (1). Work by const. force W = FΔx (F,Δx in same direction).  Net (total) work: W net = ∑FΔx (2). (N’s 2 nd Law in energy form!) Or using N’s 2 nd Law: W net = maΔx (3). ∑F is constant Acceleration a is constant  Ch. 2 kinematic equation: (v f ) 2 = (v i ) 2 + 2aΔx  a = [(v f ) 2 - (v i ) 2 ]/(2Δx) (4) Combine (4) & (3): W net = (½)m[(v f ) 2 - (v i ) 2 ] (5) xixi xfxf

Summary: Net work done by a constant net force in accelerating an object of mass m from v i to v f is: W net = (½)m(v f ) 2 - (½)m(v i ) 2   K (I) DEFINITION: Kinetic Energy (K). ( Kinetic = “motion”) K  (½)mv 2 (units are Joules, J) WORK-KINETIC ENERGY THEOREM W net =  K = K f - K i (  = “change in”) NOTE: The Work-KE Theorem (I) is 100% equivalent to N’s 2 nd Law. IT IS Newton’s 2 nd Law in work & energy language! We’ve shown this for a 1d constant net force. However, it is valid in general!

Net work on an object = Change in KE. W net =  K  (½)[m(v f ) 2 - m(v i ) 2 ] Work-Kinetic Energy Theorem –Note: W net = work done by the net (total) force. –W net is a scalar. –W net can be positive or negative (because  KE can be both + & -) –Units are Joules for both work & K.

Moving hammer can do work on nail. For hammer: W h =  K h = -Fd = 0 – (½)m h (v h ) 2 For nail: W n =  K n = Fd = (½)m n (v n ) 2 - 0

Examples v i = 20 m/s v f = 30 m/s m = 1000 kg Conceptual v i = 60 km/h v f = 0 Δx = 20 m v i = 120 km/h v f = 0 Δx = ??

Example 7.6 A block, mass m = 6 kg, is pulled from rest (v i = 0) to the right by a constant horizontal force F = 12 N. After it has been pulled for Δx = 3 m, find it’s final speed v f. Work-Kinetic Energy Theorem W net =  K  (½)[m(v f ) 2 - m(v i ) 2 ] (1) If F = 12 N is the only horizontal force, we have W net = FΔx (2) Combine (1) & (2): FΔx = (½)[m(v f ) 2 - 0] Solve for v f : (v f ) 2 = [2 Δx/m] (v f ) = [2Δx/m] ½ = 3.5 m/s

Conceptual Example 7.7