1. Work and Energy

2. Work Work = F∙s = mg ∙ (h 2 – h 1 ) = mgh 2 – mgh 1 = Δ PE = change in potential energy a = (v 2 – v 1 )/t s = ½(v 2 +v 1 )∙t as = ½(v 2 ²-v 1 ²) mas = ½m(v 2 ²-v 1 ²) F∙s = ½∙mv 2 ² - ½∙mv 1 ² = ΔKE = change in kinetic energy Work transfers energy What happens if the force is not constant?

3. Force that Varies W = F 1 ∙s 1 + F 2 ∙s 2 + F 3 ∙s 3 +... distance = s = ∆x W = F 1 ∙∆x 1 + F 2 ∙∆x 2 + F 3 ∙∆x 3 +... If Force depends on position: F = F(x) F is a function of x W = ∫ F(x) dx

4. Hooke’s Law F = k∙x W = ∫ F(x) dx = ∫ kx∙dx = ½kx² If a spring has a Hooke’s constant of 4 Newtons/meter, what work is needed to stretch it from equlibrium (x=0) to x = 2 meters?

5. Gravitational Energy Here at the surface, F = mg (constant) Over distances which are comparable to the earth’s radius: F = k / r² Integrate: W = ∫ F(x) dx = ∫ (k/r²)∙dr = -k/r

6. Gravitational Energy What energy is needed to lift a mass m from the surface of the earth (r 0 = 6,378,000 m) to a distance r 1 from the center of the earth? Integrate: W = ∫ F(x) dx = ∫ k/r²∙dr = -k/r Evaluate at two limits W = −k/r 1 + k/r 0 = k(1/r 0 −1/r 1 ) At the surface, solve for k: k = F∙r² = mg∙r 0 2 W = k(1/r 0 −1/r 1 ) = mg∙r 0 2 ∙(1/r 0 −1/r 1 ) = mgr 0 (1 − r 0 /r 1 ) mgr 0 is called the “escape energy” What velocity does this correspond to?

7. Escape Velocity Escape energy = mgr 0 = ½mv² v² = 2g/r 0 = 2 ∙ 9.8 m/s² ∙ 6,378,000 m = 125,000,000 m²/s² v = 11200 m/s = 11.2 km/s = 6.95 miles/sec

8. Suppose Acceleration is not Constant a = (v 2 – v 1 )/t s = ½(v 2 +v 1 )∙t a = a(t) v = v(t) a = dv/dt dx/dt = v a ∙ dx/dt = v ∙ dv/dt (dot product) ma ∙ dx/dt = m v∙dv/dt F ∙ dx/dt = md(½v²)/dt W = ∫ F(x) ∙ dx = ∫ ½md(v²) = ½mv 2 ² - ½mv 1 ²