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6) When released, a 100-g block slides down the path shown below, reaching the bottom with a speed of 4.00 m/s. How much work
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given m=100g m=0.1 kg at initial blolck is, in (vp = 4 m/ h=20 We know that - work done by - (change in kinetic) all the forcat initial KE:= Kinetic energy of block IKE; = 1 m vi Put eqn. the value of wg, KE, KE; in above We thg = KES-KER Wi & mgh =

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