Recent content by Fitz Watson

I've used $$mg \cdot cos\frac{\pi}{3} $$ in the solution. As for the normal force, it's not present because I've considered the instant at which the block has just left the loop

$$mg(0.45) = mg(R + R \cdot cos(\frac{π}{3})) + \frac{1}{2}mv^2$$ $$v^2 = g(0.9 - 3R)$$ The centripetal acceleration during the "flying through air" will be given by gravity $$mg \cdot cos(\frac{\pi}{3}) = \frac{mv^2}{r}$$ $$R = \frac{1.8}{5}$$ But my book says $$ R = \frac{1}{5}$$

Is this formula wrong? Bulk Modulus = Pressure/Volume Strain? I haven't gone into the differential form of it yet, so that's why I'm using this one

Today, while studying about bulk Modulus, I encountered a doubt. Please consider this thought experiment. I'm considering Caesium as an example as it seems to have a quite low Bulk Modulus (comparatively) of 1.6 GPa. Let's say I apply a pressure of X GPa. Volume change ratio can be given by...

Now that I rethink, will it be like after forces becoming equal, block will go further down due to inertia?

But when it has reached maximum extension, doesn't it mean that now the spring force is equal to weight + external force applied. And that's why the block can't go further down

Homework Statement A block is suspended by an ideal spring of the force constant K. If the block is pulled down by applying a constant force F and if maximum displacement of the block from its initial position of rest is X then, find the value of X. Homework Equations mg + F = XK + K(mg/K)...