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Figure8-5a shows two industrial spies sliding an initially stationary 225 kg floor safe a displacement vec(d) of magnitude 8.50 m. The push vec(F)_(1) of spy001 is 12.0 N at an angle of 30.0^(@) downward from the horizontal, the pull vec(F)_(2) of spy 002 is 10.0 N at 40.0^(@) above the horizontal. The magnitudes and directions of these forces do not change as the safe moves, and the floor and safe make frictionless contact. The safe is initially stationary. What is its speed v_(f) at the end of the 8.50 m displacement? |
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Answer» Solution :KEY IDEA The speed of the safe changes because its kinetic energy is changed when energy is transferred to it by `VEC(F)_(1)` and `vec(F)_(2)`. CALCULATIONS: We relate the speed to the work done by combining Eqs. 8-10 ( the work-kinetic energy theorem) and 8-1 (the definition of kinetic energy): `W=K_(f)-v_(i)=1/2 mv_(f)^(2) - 1/2 mv_(i)^(2)`. The initial speed `v_(i)` is zero, and we now know that the work done is 153.4 J. Solving for `v_(f)` and then substituting known data, we find that `v_(f)=sqrt((2W)/(m))=sqrt((2(153.4J))/(225g))` `=1.17m//s`. |
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