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

Wnet = ∑FΔx (2). (N’s 2nd Law in energy form!) Object undergoes displacement Δr = Δx i (Δx= xf - xi) & velocity change (Δv= vf -vi) under action of const. net force ∑F figure  Text derivation Calculus not needed! Instead, Newton’s 2nd Law ∑F = ma (1). Work by const. force W = FΔx (F,Δx in same direction).  Net (total) work: Wnet = ∑FΔx (2). (N’s 2nd Law in energy form!) Or using N’s 2nd Law: Wnet = maΔx (3). ∑F is constant Acceleration a is constant  Ch. 2 kinematic equation: (vf)2 = (vi)2 + 2aΔx  a = [(vf)2 - (vi)2]/(2Δx) (4) Combine (4) & (3): Wnet = (½)m[(vf)2 - (vi)2] (5) xi xf Figure 7.12: An object undergoing a displacement Δr = Δx î and a change in velocity under the action of a constant net force ∑F.

Wnet = (½)m(vf)2 - (½)m(vi)2  K (I) WORK-KINETIC ENERGY THEOREM Summary: Net work done by a constant net force in accelerating an object of mass m from vi to vf is: Wnet = (½)m(vf)2 - (½)m(vi)2  K (I) DEFINITION: Kinetic Energy (K). (Kinetic = “motion”) K  (½)mv2 (units are Joules, J) WORK-KINETIC ENERGY THEOREM Wnet = K = Kf - Ki ( = “change in”) NOTE: The Work-KE Theorem (I) is 100% equivalent to N’s 2nd Law. IT IS Newton’s 2nd Law in work & energy language! We’ve shown this for a 1d constant net force. However, it is valid in general!

Work-Kinetic Energy Theorem Net work on an object = Change in KE. Wnet = K  (½)[m(vf)2 - m(vi)2] Work-Kinetic Energy Theorem Note: Wnet = work done by the net (total) force. Wnet is a scalar. Wnet 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: Wh = Kh = -Fd = 0 – (½)mh(vh)2 For nail: Wn = Kn = Fd = (½)mn(vn)2 - 0

Examples Conceptual vi = 20 m/s vf = 30 m/s m = 1000 kg vi = 60 km/h Δx = 20 m vi = 120 km/h vf = 0 Δx = ??

Example 7.6 Work-Kinetic Energy Theorem A block, mass m = 6 kg, is pulled from rest (vi = 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 vf. Work-Kinetic Energy Theorem Wnet = K  (½)[m(vf)2 - m(vi)2] (1) If F = 12 N is the only horizontal force, we have Wnet = FΔx (2) Combine (1) & (2): FΔx = (½)[m(vf)2 - 0] Solve for vf: (vf)2 = [2Δx/m] (vf) = [2Δx/m]½ = 3.5 m/s Figure 7.13: (Example 7.6) A block pulled to the right on a frictionless surface by a constant horizontal force.

Conceptual Example 7.7 Figure 7.14: (Conceptual Example 7.7) A refrigerator attached to a frictionless, wheeled hand truck is moved up a ramp at constant speed.