Dropping a rock

If you drop a 1-kg rock from a height of 1 m, how many joules of KE does it have on impact with the ground? (Assume that any energy transformed into heat by air friction is negligible.)

If we choose the y axis to point up, then F_y is negative, and equals -(1 kg)(g) = -9.8 N. A decrease in y is represented by a negative value of Δy, Δy = -1 m, so the change in potential energy is -(-9.8 N)(-1 m) ~ -10 J. (The proof that newtons multiplied by meters give units of joules is left as a homework problem.) Conservation of energy says that the loss of this amount of PE must be accompanied by a corresponding increase in KE of 10 J.