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EXAMPLE: BLOCK AND TACKLE
  1. ⎲⎳Fx = ma → T − 2  =  (10)/(32.2)a       ⎲⎳Fy = ma →  − T + 5  =  (5)/(32.2)a       3 = (15)/(32.2)a  →  a = 6.44(f)/(s2)       ÿ  =  6.44       ẏ  =  6.44t + 0       y  =  3.22t2 + 0 → t = .788sec
  2. ẏ = 6.44(.788)  =  5.08(ft)/(s)       or *1cmW  =  ΔT       T(2) − 2(2) + 5(2) − T(2)  =  (1)/(2)mv22 − (1)/(2)mV21 *1cm[(1)/(2)mV21 = 0]       10 − 4  =  (7.5)/(g)V2       V  =  5.08(ft)/(s)
  3. figure images/Quiz2_fbd_w_spring-1.jpg
    Figure 4 Free body diagram when a spring is added.
    With the addition of the spring (see Fig. 4↑), Fs should balance T + Ff. Now T=mg=5
    ⎲⎳Fx = 0  =   − kδ + 2 + 5 *1cm[δ = 2]       7  =  k*2       k  =  (7)/(2) = 3.5      
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