posted by Amber on .
The figure below shows the speed of a person's body as he does a chin-up. Assume the motion is vertical and the mass of the person's body is 73.4 kg. Determine the force exerted by the chin-up bar on his body at the following times. t=0s, speed= 0cm per second; t= .5s, speed= 15cm per second; t=1.1s, speed =28cm per second; t=1.6s, speed=6cm per second.
I have no idea how to even begin to approach this problem. I am given a mass, time, and speed but Im asked to find a force. If anyone could please help me I would greatly appreciate it.
The force of the person on the chin-up bar, or vice versa, depends upon the weight Mg and the acceleration of the person, not his or her velocity.
F = M (g + a) will give you the force on the bar.
You need to use the figure (which was not provided in the post) to get the instantaneous acceleration somehow.
Oh okay, thank you. The only thing included was a graph giving the points t=0s, speed= 0cm per second; t= .5s, speed= 15cm per second; t=1.1s, speed =28cm per second; t=1.6s, speed=6cm per second.
Sorry for so many posts, but you would not use delta v/ delta t to find acceleration. correct? on the graph it shows that at 1.1 second the graph changes directions therefore the slope of the tangent line at that point is equal to zero. I plugged it into the formula you gave me and I got that correct. How should I get the other instantaneous accelerations from the graph? Thank you so much for your help.