Question

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

Answer 1

Use N - mg = ma (upward positive). So N = m(g + a) = 60.0(9.80 + 1.80) = 60.0(11.60) = 696 N.

The floor exerts a 696-N upward normal force on the boy.