Recent content by method_man

As far as I know it should be. Try to google and find something about Pascal's Vases or hydrostatic paradox.

Hello, Try to draw your FBD a little bit precise. You've got three forces acting on your block: ma, G=mg and N.

You have a similar thread https://www.physicsforums.com/showthread.php?t=259862

What is the work equal to? Friction and potential energy are not doing the same job as in a).

F = -kx = -k[\sqrt{x^{2}+L^{2}} - L]--->Force of one spring \sqrt{x^{2}+L^{2}}-->stretched length of a spring \frac{x}{\sqrt{x^{2}+L^{2}}}--->cosine between x and stretched length -k[\sqrt{x^{2}+L^{2}} - L] * \frac{x}{\sqrt{x^{2}+L^{2}}} i.e. -kx[\sqrt{x^{2}+L^{2}} - L] *...

Until now, you have done a good job. Now, look at the solution. The result is a vector force. It's a component of a red force in my photo, i.e. that is a horizontal component of a spring force.

Do you know anything about statics? You have two unknowns and you need two equations. It would probbably help you if you would draw a picture. You need to make a free body diagram first.

I would say that you also need to calculate the work done. Because the piston is "free to move". If the piston has moved, there has been a change in the volume. If there is a change in the volume, there is work done.

Draw a v-t diagram and calculate a surface under the lines.

I think you made a mistake by determening an angle.

I forgot to erase that when I quoted. My bad.

I would say that this is the right equation:T2R sin - T1R sin = I alpha . The disk is rotating clockwise, so the torque "I alpha" is directed counterclockwise. It's similar to the reactional force m*a which is directed opposite from the acceleration.

How much is the spring compressed? http://www.physics247.com/physics-tutorial/spring-potential-energy.shtml"

Yes. You have a nice example http://www.fsas.upm.edu.my/~zainalas/PHY2001/2006/example4.pdf" on page 6.

Perhaps you didn't detail your answer enough. The question is, what do you have to do to stop your car? Change it's kinetic energy. So, work done is equal to the difference between final and initial kinetic energy: W=EKF-EKI