Julie's Jump: Solving Kinetic Energy & Internal Energy Problems

[Kreamer]

Homework Statement

Julie has a mass of 60 kg. In an experiment she crouched down, then jumped straight up. Her lab partners
measured the height of her center of mass above the floor at three instants: 1) 0.43 m when crouched
down; 2) 1.02 m just as her feet were leaving the floor; 3) the height at the top of the jump. The height at
the top of the jump is not given here, but her lab partners used this to calculate that Julie’s translational
kinetic energy as her feet left the floor was 88 J.

(a) Using the “real” system, find the change in Julie’s internal energy from the crouched position to just
as her feet leave the floor.

(b) Using the point particle system, find the magnitude of the force that the floor exerted on the bottom
of Julie’s feet while she was in contact with the floor. You may assume that this force is approximately
constant while her feet touch the floor.

(c) Using the point particle system, find the maximum height (at the top of the jump) that Julie’s center
of mass reaches above the floor.


The attempt at a solution
I have tried so many things in an attempt at this question but none seem to work. It is a practice test so I know the answers but nothing I do reflects the answers. I have a test soon so if anyone could walk me through the process of solving this I would be very grateful
Answers:
a)-435 J
b) 737 N
c) 1.17 m

 

[anathema]

Hello, Julie! I would be happy to help you with this problem. Let's break it down step by step and use some equations to solve for the answers.

(a) To find the change in Julie's internal energy, we first need to understand what internal energy is. Internal energy is the total energy stored in a system, including the kinetic energy of its particles and the potential energy between them. In this case, Julie's internal energy will change as she goes from a crouched position to jumping in the air.

We can use the equation for change in internal energy (ΔU) to solve for this. It is given by ΔU = U2 - U1, where U2 is the final internal energy and U1 is the initial internal energy. In this case, U1 is the internal energy when Julie is crouched down and U2 is the internal energy just as her feet leave the floor.

We can also use the equation for translational kinetic energy (K) to find U2. It is given by K = 1/2mv^2, where m is the mass and v is the velocity. In this case, we know that Julie's kinetic energy just as her feet leave the floor is 88 J. So we can write:

88 J = 1/2(60 kg)v^2

Solving for v, we get v = √(88 J x 2 / 60 kg) = 2.98 m/s

Now, we can use the equation for internal energy to find the change in Julie's internal energy:

ΔU = U2 - U1 = 1/2mv^2 - 1/2mv^2 = 1/2(60 kg)(2.98 m/s)^2 - 1/2(60 kg)(0.43 m/s)^2 = 435 J

Therefore, the change in Julie's internal energy is -435 J (note the negative sign, indicating a decrease in internal energy).

(b) To find the magnitude of the force exerted by the floor on Julie's feet, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) times acceleration (a). In this case, the force is the normal force exerted by the floor on Julie's feet, the mass is her mass of 60 kg, and the acceleration is the change