What is the acceleration when the cord has been stretched?

[mahtab_lovelygirl]

Please help me :(

i have lots of question about physic...
i hope u guys can help me. i really appreciate :)

1)
a) now that i am 50 and semi-mature. you'll never catch me bungee jumping. but let's say someone my mass(80kg) and height (1.83m) did and they used a 20m bungee cord with a k value of 500 n/m. what is the minimum height that the lunatic would need in order that he/she won't get killed?

b) if the same person was gently lowered down and the bungee cord came to equilibruim, far far would the cord stretch?

c)what is the person's speed when they have stretched the cord 8m? consider energy transfor motion.

d) what is the acceleration when the cord has been stretched?



2) an electron moving threough an electric field of 475 v/m and a magnetic field of oil experiences no force. if the electron's direction and the directions of the electric and magnetic fields are all mutually perpendicular, what is the speed of the electron? start with a sketch :P

i really need them as soon as possible for my final exam:(

please help me :!)

 

[pete worthington]

Welcome to Physics Forum...We can help you, but you must do the work...
1a) Conservation of energy. Initial potential-gravitational energy will transform into final potential-spring energy. You have all the variables except height. As far as the persons height, that depends on how/where the person jumps from. The length of the cord must also be taken into account. Draw a picture and label positions.
1b) Hookes Law. That equation should be easy to find.
1c) Go back to your drawing. At this position you have potential-gravitational, kinetic and potential-spring. The unknown is velocity.
1d) Hookes Law and Newton's Second Law combined.

 

[thepatientmental]

1) a) To determine the minimum height needed for the person to not get killed, we can use the formula for elastic potential energy: PE = 1/2kx^2, where k is the spring constant (in this case, the bungee cord's k value) and x is the distance the cord is stretched. We can set the elastic potential energy equal to the person's gravitational potential energy at the minimum height, which is mgh, where m is the person's mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the minimum height. So, we have 1/2kx^2 = mgh. Plugging in the given values, we get 1/2(500)(20)^2 = (80)(9.8)h. Solving for h, we get a minimum height of approximately 20.4 meters.

b) If the person is gently lowered down and the bungee cord reaches equilibrium, it means that the forces acting on the person are balanced. This means that the weight of the person (mg) is equal to the spring force of the bungee cord (kx). So, we can set these two forces equal to each other and solve for x, the distance the cord stretches. We get mg = kx, or x = mg/k. Plugging in the given values, we get x = (80)(9.8)/(500) = 1.568 meters.

c) To find the person's speed when the cord has been stretched 8m, we can use the equation for kinetic energy: KE = 1/2mv^2, where m is the person's mass and v is their speed. We can set the kinetic energy equal to the elastic potential energy at 8m, which is 1/2kx^2. So, we have 1/2mv^2 = 1/2k(8)^2. Solving for v, we get a speed of approximately 13.14 m/s.

d) The acceleration when the cord has been stretched is calculated using Newton's Second Law: F = ma. The net force acting on the person is the difference between their weight (mg) and the spring force of the bungee cord (kx). So, we have F = mg - kx. Plugging in the given values, we get F