We can't call it Fluber. Assume a certain liquid, with density 1230 kg / m^3 , exerts no frictio

We can't call it Fluber. Assume a certain liquid, with...

We can't call it Fluber. Assume a certain liquid, with density 1230 $\mathrm{kg} / \mathrm{m}^{3}$ , exerts no friction force on spherical objects. A ball of mass 2.10 $\mathrm{kg}$ and radius 9.00 $\mathrm{cm}$ is dropped from rest into a deep tank of this liquid from a height of 3.30 $\mathrm{m}$ above the surface. (a) Find the speed at which the ball enters the liquid. (b) What two forces are exerted on the ball as it moves through the liquid? (e) How does the time interval $\Delta t_{\text { down }}$ , during which the ball moves from the surface down to its lowest point, compare with the time interval $\Delta l_{\text { up }}$ for the return trip between the same two points? (f) What If? Now modify the model to suppose the liquid exerts a small friction force on the ball, opposite in direction to its motion. In this case, how do the time intervals $\Delta l_{\text { down }}$ and $\Delta t_{\text { up }}$ compare? Explain your answer with a conceptual argument rather than a numerical calculation.

Key Concepts

Conservation of Energy

This principle states that the total energy of an isolated system remains constant, even though energy can transform from one form to another. In the context of falling objects, gravitational potential energy is converted to kinetic energy, allowing calculation of the speed of an object immediately before impact with another medium, assuming negligible energy losses.

Buoyant Force

Buoyancy is the upward force exerted by a fluid on an immersed object, as described by Archimedes' principle. This force is equal to the weight of the displaced fluid and is key in determining the net force acting on objects when they are submerged in a liquid.

Resistive (Drag) Force in Fluids

Moving objects in a fluid can experience a resistive force, commonly known as drag, which opposes the direction of motion. This force depends on factors such as the fluid’s density, the object's speed, and its cross-sectional area, influencing the acceleration and deceleration during motion through the fluid.

Newton's Second Law of Motion

This law, expressed as F = ma, relates the net force acting on an object to its mass and the acceleration produced. It provides the framework for analyzing the multiple forces acting on an object, such as gravity, buoyancy, and friction (or drag) forces in fluid environments.

Time Symmetry in Conservative Systems

In the absence of non-conservative forces like friction, the motion of an object under a conservative force field is time-reversible. This means that the time required to traverse a path in one direction is equal to the time required to return along the same path, a concept that is disrupted when resistive forces are present.

Transcript

00:03 So here for our solution for part a, so we have to apply v raised to the power of 2 s minus u raised to the power of 2 which is equals to 2 negative g times y.

00:21 So v s2 this would be equal to negative 2 times 9 .8 meter per second squared times negative 3 .30 meters.

00:33 So we have v s which is equals to 8 .0424 meter per seconds.

00:42 For part b, the two forces exerted on the ball, so the the two forces the two forces exerted on the ball as it moves through the liquid are the weight are the weight of the ball acting vertically downward and buoyant force acting on the ball which is upwards.

01:34 Now for part c, so we have 4 over 3 pi r raised to the power of 3 which is equals to 4 divided by 3 pi that is 0 .09 meter raised to the power of 3.

01:52 So we have equal to 0 .003, so we have 0 .00305 cubic meter.

02:07 And here we need to identify the net force, so we have fb minus mg.

02:18 So we have equal to 36 .79 newton minus 2 .1 kilogram.

02:26 So 2 .1 kilogram times 9 .8 meter per seconds raised to the power of 2.

02:32 So this would be equal to 36 .79 newton minus 20 .88 newton.

02:41 So we have equal to 16 .21 newton.

02:47 Next one we need to apply to identify the distance traveled.

02:52 So we have s is equals to 8 .0424 meter per seconds raised to the power of 2 times 2 .10 kilogram divided by 2 times 1, 6 .21 newton.

03:11 So we have equal to 4 .19 meters...