Philip P. answered • 09/15/16

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You use both. When the weight is being held above the floor, it has potential energy (E

_{p}): E

_{p}= mgh = (150 kg)(9.8 m/s^{2})(2 m) = 2940 joulesWhen the weight is dropped, the potential energy is converted into kinetic energy (E

_{k}): E

_{k}= (1/2)mv^{2} 2940 joules = (1/2)(150 kg)v

^{2}Solve for v. The units are m/s.

Shih T.

09/18/16