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.

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.