Solve Mech. Energy Problem: Homework Statement & Equations

[lgmavs41]

Just need some direction on the problem.

Homework Statement

A poorly designed playground slide begins with a straight section and ends with a circular arc. A child starts at point P and slides down both sections of the slide. At some point on the circular arc, the normal force goes to zero and the child loses contact with the ramp. Assuming the forces of friction are negligible, at what height from the ground will the child become airborne.

the height from point p to the ground is 9 m. The radius of the arc is 7.2 m.

Homework Equations

w=kf-ki+uf-ui; ki=0, ui=mg(9)

The Attempt at a Solution

well, i figure out the speed in which the particle will be when it hits the circular arc, where uf=mg(7.2). Now how do i figure out how high it will be from the ground when the child will become airborne? I think I need to use F=mv^2/r somewhere in the equation to figure out the force needed for the particle to stay in the circular path and not go flying off. But after that, I'm kind of lost.

Thanks for the help.

 

[lgmavs41]

BTW, the child touches the circular arc at the top of the circle which covers about a quarter of the circumference...well, picture the child sliding straight down then suddenly touches the top of the circle, which slides down in an arc. Sorry, i don't have the pic.

 

[m2003]

Hi there,

Thank you for providing the homework statement and equations for this problem. It seems like you have made some progress in solving it already. Let's break down the steps you have taken so far and see if we can provide some direction for you to continue.

First, you correctly identified that the child's initial kinetic energy (ki) is equal to 0, since they start at rest at point P. The potential energy at point P (ui) is equal to mg(9), where m is the mass of the child, g is the acceleration due to gravity, and 9 is the height from point P to the ground.

Next, you found the final velocity (uf) of the child when they reach the circular arc, using the equation uf=mg(7.2). This is the point where the normal force goes to zero and the child becomes airborne.

To find the height from the ground where the child becomes airborne, we can use the conservation of mechanical energy equation, w=kf-ki+uf-ui, where w is the work done, kf is the final kinetic energy, and ui is the initial potential energy. Since we know that ki=0 and uf=mg(7.2), we can rewrite the equation as w=mg(7.2)-ui. The work done, w, is equal to the change in potential energy, so we can rewrite the equation again as mg(7.2)-ui=mg(9)-ui. Solving for ui, we get ui=mg(7.2)-mg(9)=mg(2.8). This means that the child has lost mg(2.8) of potential energy when they reach the circular arc.

To find the height from the ground where the child becomes airborne, we can use the equation ui=mg(2.8). Solving for h, we get h=2.8 meters.

So, the child will become airborne when they are at a height of 2.8 meters from the ground. I hope this helps guide you in solving this problem. Remember to always use the conservation of mechanical energy equation and to consider all forms of energy (kinetic, potential, and work) when solving these types of problems. Good luck!