How Does a Skateboarder's Movement Affect Work Done by Forces in a Half-Pipe?

[Tmtamrak]

A skateboarder with his board can be modeled as a particle of mass 73.0 kg, located at his center of mass (which we will study in a later chapter). As shown in the figure below, the skateboarder starts from rest in a crouching position at one lip of a half-pipe (point A). The half-pipe is a dry water channel, forming one half of a cylinder of radius 6.20 m with its axis horizontal. On his descent, the skateboarder moves without friction so that his center of mass moves through one quarter of a circle of radius 5.70 m. Immediately after passing point B, he stands up and raises his arms, lifting his center of mass from 0.500 m to 0.880 m above the concrete (point C). To account for the conversion of chemical into mechanical energy, model his legs as doing work by pushing him vertically up, with a constant force equal to the normal force nB, over a distance of 0.380 m. (You will be able to solve this problem with a more accurate model described in a later chapter.)
d. What is the work done on the skateboarder's body in this process




(mv^2)/r; W=F delta r



I calculated the normal force to be 2146.26N which is mg + (mv^2)/r. However, knowing the work equation, what force do I use to calculate this work? Also, I have the angle as 0.. is that correct? My answer seems to be wrong.
2146.26 x 0.38m

 

[anathema]

= 815.3J

it is important to consider all the forces acting on the skateboarder and the work done by each force. In this scenario, the skateboarder experiences three main forces: gravity, normal force, and the force from his legs pushing him up.

First, let's calculate the work done by gravity. Since the skateboarder is moving in a circular path, the force of gravity does not do any work since it is perpendicular to the displacement of the skateboarder's center of mass. Therefore, the work done by gravity is zero.

Next, let's calculate the work done by the normal force. As you correctly calculated, the normal force is equal to the weight of the skateboarder plus the centripetal force required for him to move in a circular path. This centripetal force is provided by the normal force from the half-pipe. So, the work done by the normal force is:

W = nB * 0.38m = (mg + (mv^2)/r) * 0.38m = (73.0kg * 9.8m/s^2 + (73.0kg * (5.70m/s)^2)/6.20m) * 0.38m = 2146.3J

Lastly, let's calculate the work done by the force from the skateboarder's legs pushing him up. This force is equal to the normal force at point C, since the skateboarder is standing up and his center of mass is at the same height as point C. So, the work done by the force from his legs is:

W = nC * 0.380m = 2146.3J

Therefore, the total work done on the skateboarder's body is the sum of these three works:

W = 0J + 2146.3J + 2146.3J = 4292.6J

To answer your question about the angle, since the skateboarder is moving in a circular path, the angle between the force of gravity and the displacement of the skateboarder's center of mass is constantly changing. Therefore, it is not necessary to include it in the calculation.

I hope this helps clarify the work done in this scenario. Keep in mind that this is a simplified model and a more accurate model would take into account the conversion of chemical energy into mechanical energy by the skateboarder's legs, as