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.

To solve this problem, we need to apply Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration. In this case, the acceleration is equivalent to the rate of change of velocity, which we can calculate using the given information.

To find the force exerted by the chin-up bar on the person's body, we can follow these steps for each given time:

Step 1: Convert speed from cm per second to m per second.
Given speeds are:
- 0 cm/s
- 15 cm/s
- 28 cm/s
- 6 cm/s

To convert cm/s to m/s, divide the given speeds by 100:
- 0 cm/s ÷ 100 = 0 m/s
- 15 cm/s ÷ 100 = 0.15 m/s
- 28 cm/s ÷ 100 = 0.28 m/s
- 6 cm/s ÷ 100 = 0.06 m/s

Step 2: Calculate the acceleration using the equation: acceleration = (final speed - initial speed) / time interval.
Given time intervals are:
- t = 0s
- t = 0.5s
- t = 1.1s
- t = 1.6s

Calculate acceleration for each time interval:
- At t = 0s, since the speed is 0, the acceleration is also 0.
- At t = 0.5s: acceleration = (0.15 m/s - 0 m/s) / 0.5s = 0.15 m/s ÷ 0.5s = 0.3 m/s²
- At t = 1.1s: acceleration = (0.28 m/s - 0.15 m/s) / (1.1s - 0.5s) = 0.13 m/s ÷ 0.6s ≈ 0.22 m/s²
- At t = 1.6s: acceleration = (0.06 m/s - 0.28 m/s) / (1.6s - 1.1s) = -0.22 m/s ÷ 0.5s = -0.44 m/s²

Step 3: Calculate the force using Newton's second law of motion: force = mass × acceleration.
The given mass is 73.4 kg.

For each time interval, calculate the force exerted by the chin-up bar:
- At t = 0s, force = 73.4 kg × 0 m/s² = 0 N
- At t = 0.5s, force = 73.4 kg × 0.3 m/s² = 22.02 N
- At t = 1.1s, force = 73.4 kg × 0.22 m/s² = 16.148 N
- At t = 1.6s, force = 73.4 kg × (-0.44 m/s²) = -32.296 N

Note: The negative force at t = 1.6s indicates that the chin-up bar is exerting a downward force, which opposes the upward motion of the person's body.

Therefore, the force exerted by the chin-up bar on the person's body is:
- At t = 0s: 0 N
- At t = 0.5s: 22.02 N
- At t = 1.1s: 16.148 N
- At t = 1.6s: -32.296 N