Question

A 60.0-kg boy rides in an elevator going up while slowing down at -1.80 m/s2. What is the magnitude of the normal force exerted on the boy by the floor work it out

Answer 1

Take upward as positive. Sum forces: N - mg = ma, so N = m(g + a). With m = 60.0 kg, g = 9.80 m/s², a = -1.80 m/s²:

N = 60.0(9.80 - 1.80) = 60.0(8.00) = 480 N.

Magnitude of the normal force = 4.80 × 10^2 N.