There is a graph of the distance and potential energy. When the problem states that it is 4cm from its equilibrium position, is it referring to the x or y component?
Kinetic and Potential Energy!?!
This problem is totally confusing me..can someone please explain it...:confused:
A 3.0 kg object subject to a restoring force F is undergoing simple harmonic motion with a small amplitude. The potential energy U of the object as a function of distance x from...
The exact problem is:
A massless spring is between a 1-kilogram mass and a 3-kilogram mass, but it is not attached to either mass. Both masses are on a horizontal frictionless table. In an experiment, the 1-kilogram mass is held in place, and the spring is compressed by pushing on the...
Ok...Here is the problem.
A block of mass 4 kilograms, which has an initial speed of 6 meters per second at time t=0, slides on a horizontal surface.
(a) Calculate the work W that must be done on the block to bring it to rest.
For this problem, I used the formula W= 1/2mvf ^2– 1/2mvi ^2...
I really mix up and/or combine velocity and acceleration. The change from constant acceleration to constant velocity really confused me, because I know that acceleration can remain constant.
So, if I use 5m/s^2 and solve for velocity, I get 10m/s, then when I solve for the time, I get 10s. Is that right, because I got really confused when it mentioned constant acceleration and velocity.
Ok, I have this problem...
The first 10 meters of a 100 meter dash are covered in 2 seconds by a sprinter who starts from rest and acceleraties with a constant acceleration. The remaining 90 meters are run with the same velocity the stprinter had after 2 seconds.
a) Determin the sprinter's...
does anyone know an easier way to determine which formula is used for calculating moment of inertia? i have all the formulas and descriptions, but i have trouble figuring out which formula to use based on what is given in the problem. any ideas thatll help me?
I solved for the final velocity, and I got 10.51m/s. Then I used the conservation of momentum equation, and for the velocity of the croquet ball, I got 4.95m/s. So, to find the distance from the collision point, would I just use this kinematics equation, x=1/2Vft?
The answer that I got is...
To find the velocity of the croquet ball, is that even possible, without the time? Could I use the .80 seconds as a point in time and base my info off of the collision?
I swear, this is the last problem that I have.
Here it goes...
A pool ball having a mass of .65 kg experiences an elastic collision with a croquet ball having a mass of 1.38 kg. The pool ball had been rolling for 4.3 seconds, having covered a distance of 22.6 meters during that time. The...
Ok, let me check what I did. I just wanted to make sure, because it really didn't sound very realistic to me. Great! I redid the problem, and everything came up equivalent to each other.
Thanks for your help. It's really gonna help me on my exam tomorrow! :)
When I solved for the final velocity, I solved for potential gravitational energy, I multiplied mgh, and I got 588.6. Then I set that equal to kinetic energy, which is 1/2mv^2. I solved for velocity, and I got 6.26 m/s. From there, I took the velocity and used a kinematics equation to solve for...