Question

What is the maximum velocity of an 85.0-kg person bouncing on a bathroom scale having a force constant of (1.50 X 10^6 N/m), if the amplitude of the bounce is (0.200 cm)?

a) (1.23 m/s)
b) (2.46 m/s)
c) (3.69 m/s)
d) (4.92 m/s)

Answer 1

The maximum velocity of an 85.0-kg person bouncing on a bathroom scale with a known force constant and amplitude can be calculated using the formula for maximum velocity in simple harmonic motion. In this question, the calculated maximum velocity is approximately 0.2657 m/s, which does not match the given multiple-choice options. The maximum energy stored in the spring can also be calculated and is about 3 joules.

Step-by-step explanation:

To determine the maximum velocity of an 85.0-kg person bouncing on a bathroom scale with a force constant of 1.50×106 N/m and an amplitude of 0.200 cm, we use the relationship between the maximum velocity (vmax) and the amplitude (A) in simple harmonic motion for a spring, which is given by:

vmax = ωA

where ω is the angular frequency. The angular frequency is calculated using the mass (m) and spring constant (k) as:

ω = √(k/m)

First, we convert the amplitude from cm to meters:

A = 0.200 cm × 10-2 m/cm = 0.002 m

Now let's calculate the angular frequency:

ω = √(1.50×106 N/m / 85.0 kg) = √(17647.1 rad2/s2) ≈ 132.85 rad/s

Now we can find the maximum velocity:

vmax = 132.85 rad/s × 0.002 m ≈ 0.2657 m/s

As this value does not match any of the options given, check calculations and units for any errors.

The maximum energy stored in the spring (Emax) at the amplitude is given by:

Emax = 1/2 kA2

Emax = 1/2 × 1.50×106 N/m × (0.002 m)2 ≈ 3 J