A 1.0 kg flashlight falls to the floor. At the point during its fall when it is 0.70 m above the floor, its potential energy exactly equals its kinetic energy. How fast is it moving?
A) 45 m/s
B) 6.9 m/s
C) 14 m/s
D) 3.7 m/s
E) 9.8 m/s

Question

Quizplus · Accepted Answer

The potential energy (PE) at 0.70 m is mgh = 1.0 kg * 9.8 m/s² * 0.70 m = 6.86 J. Since PE equals kinetic energy (KE), KE = 6.86 J = 0.5 * 1.0 kg * v². Solving for v gives v = sqrt(2 * 6.86) = 3.7 m/s.

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