Question

Suppose a 78 kg person is bouncing on a bathroom scale which has a force constant of 1.5 × 106 n/m. 50% part (a) what is the maximum velocity of the person, in meters per second, if the amplitude of the bounce is 0.115 cm?

Answer 1

m = mass of the person bouncing = 78 kg

k = spring constant of the bathroom scale = 1.5 x 10⁶ N/m

v = maximum velocity of the person

A = maximum compression of the spring of scale = 0.115 cm = 0.00115 m

h = height gained at maximum speed = A = 0.00115 m

Using conservation of energy

spring potential energy = kinetic energy + gravitational potential energy

(0.5) k A² = (0.5) m v² + m g h

k A² = m v² + 2 m g h

(1.5 x 10⁶) (0.00115)² = (78) v² + 2 (78) (9.8) (0.00115)

v = 0.0538 m/s