ch07-p003 - 2 3. (a) From Table 2-1, we have v 2 = v0 + 2ax...

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3. (a) From Table 2-1, we have vv a x 2 0 2 2 =+Δ . Thus, () ( ) 2 27 1 5 2 7 0 2 2.4 10 m/s 2 3.6 10 m/s 0.035 m 2.9 10 m/s. a x =+ Δ = × + × = × (b) The initial kinetic energy is ( ) 2 22 7 7 1 3 0 11 1.67 10 kg 2.4 10 m/s 4.8 10 J. i Km v −− == × × = × The final kinetic energy is ( ) 2 7 7 1 3 1.67 10
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