Question

A woman whose body mass is 50 kg rapidly stands from a sitting position with a peak vertical acceleration of 5 m/s2 directed upwards. Relative to her body weight, what percent increase is the peak magnitude of the ground reaction force?

Answer 1

Answer:51.02

Explanation:

Given

mass of woman(m)=50 kg

acceleration (a)$=5 m/s^2$

Force acting on woman is

mg

Normal reaction(N)

Thus

N-mg=ma

N=m(g+a)

$N=50\times (9.8+5)=740$

Original Normal reaction=mg

$N_0=50\times 9.8=490 N$

% increase$=\frac{N-N_0}{N_0}$

$=\frac{740-490}{490}=0.5102$

51.02 %