Recent content by BetoG93

What is the title of the book?

I think the same. Are you sure this is the correct answer?

No, you can't have a speed greater than c. In relativity, you must use Lorentz's velocity transformation equations.

That extra force you mention is in fact the weight of the cube.

To find the difference in height, try 1.6 m - (0.18 m) sen 40°. Calculate gravitational energy and equate it to elastic potential energy.

The height h doesn't depends of inclination; it is measured from the ground. From the drawing you made, it is clear that not all gravitational potential energy goes into compressing the spring, if you take ground as a point of 0 gravitational potential energy. Try to find the difference in...

A spring is comprensed or stretched along only one axis. That said, it seems that the confusion here is what must we use as x. If 12 cm (0.12 m) is the compressed lenght, then x = 0.12 m. But if 0.12 m is the final length of the spring, x would be instead 18 cm. One question: is the spring...

Part of the kinetic energy is lost due to friction, and part is converted to gravitational potential energy. Ef - Ei = -Fd = mgh - (1/2)mv^2

Right, the rope makes an angle of 100° with the body. With this in mind, you can draw a component vector that begins from the body and goes up to the tip of the tension vector, in such way that the component is perpendicular to the body. You have then a 90°-80°-10° triangle, from which you can...

Find the function F that defines force (it is a sine or cosine, from what you told). Integrate it over the time interval from 0 to 2 seconds.

Yes, friction forces are applied at different distances from the axis. That said, the torque is greater for cilinder A, and it comes first. One question: is the lower part of cilinder B contributing to normal force? If it isn't, the problem is now solved.

I got it. The difference in the radius of cilinder A and the radius of the bits of cilinder B should account for the difference in angular speed...

But, if the inner surfaces that are in contact to the two sides (bases) of the cilinder are not frictionless, there would be a reduction of the torque... And it would also reduce kinetic energy... Anyway, ignoring that, all seems to indicate that they come at the same time.

But, if the inner surfaces that are in contact to the two sides (bases) of the cilinder are not frictionless, there would be a reduction of the torque... And they would also reduce kinetic energy... Anyway, all seems to indicate that they come at the same time.

If both cilinders have the same normal force (which is evenly distributed in the bits), then the friction force is the same, and they come to the floor at the same time. Cilinder B could be lower than A, but I think we can ignore this.