(Solved): As part of an exercise routine, a woman does a set of chin-ups. Shown is a graph of the woman's vert ...

As part of an exercise routine, a woman does a set of chin-ups. Shown is a graph of the woman's vertical speed as she performs a chin-up. If her mass is 53.5 kg, calculate the force exerted by the chin-up bar on her hands for the given times. (Enter the magnitudes in N.)

A curve of the woman's vertical speed is graphed on a coordinate plane.

(a)t = 0

(b)t = 0.5 s

(c)t = 1.1 s

(d)t = 1.6 s

N.) (a) \( t=0 \) weight. Solve for \( F_{\text {bar. }} \) Don't forget to convert the acceleration units to \( \mathrm{m} / \mathrm{s}^{2} . \mathrm{N} \) (b) \( t=0.5 \mathrm{~s} \) Note that the slope of the speed-versus-time graph at \( t=0.5 \mathrm{~s} \) is equal to the slope at \( t=0 \). Therefore, you should find that the force for part (b) is equal to the force for part (a). \( \mathrm{N} \) (c) \( t=1.1 \mathrm{~s} \) This is the value of the acceleration at \( 1.1 \mathrm{~s} \); you are looking for \( F_{\text {bar' }} \) the force from the bar. Solve Newton's second law, \( \sum F_{y}=0 \), for \( F_{\text {bar }} \). \( \mathrm{N} \) (d) \( t=1.6 \mathrm{~s} \)