Assume a certain liquid, with density 1230 kg / m^3 exerts...

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) Evaluate the magnitudes of the two forces that are exerted on the ball as it moves through the liquid. (c) Explain why the ball moves down only a limited distance into the liquid and calculate this distance. (d) With what speed will the ball pop up out of 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 $\Delta t_{\text {up }}$ 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 $\Delta t_{\text {down }}$ and $\Delta t_{\text {up }}$ compare? Explain your answer with a conceptual argument rather than a numerical calculation.

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) Evaluate the magnitudes of the two forces that are exerted on the ball as it moves through the liquid. (c) Explain why the ball moves down only a limited distance into the liquid and calculate this distance.
(d) With what speed will the ball pop up out of 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 $\Delta t_{\text {up }}$ 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 $\Delta t_{\text {down }}$ and $\Delta t_{\text {up }}$ compare? Explain your answer with a conceptual argument rather than a numerical calculation.

Key Concepts

Effects of Dissipative Forces

When non-conservative forces such as friction or drag are present, they always act opposite to the direction of motion and dissipate energy from the system. This energy loss results in different accelerations and time intervals for upward and downward motion, breaking the time symmetry seen in purely conservative systems.

Symmetry in Conservative Systems

In a system where only conservative forces (like gravity and buoyancy) are present, the motion is time-reversible. This means that the time intervals for an object moving downward over a certain distance are equal to the time intervals for moving upward over the same path, reflecting the conservation of energy.

Kinematics of Uniformly Accelerated Motion

The study of motion under constant acceleration uses kinematic equations that relate displacement, initial and final velocities, time, and acceleration. This framework is essential for calculating parameters like the speed upon entering a fluid after free fall and the distance traveled under constant net acceleration.

Net Force and Equations of Motion

Newton’s second law relates the net force acting on an object to its acceleration. In the context of objects moving through a fluid, this net force is the sum of gravitational force, buoyant force, and any frictional (or drag) forces. Formulating the net force enables the use of kinematic equations to determine acceleration, speed, displacement, and the resulting motion of the object.

Buoyant Force (Archimedes' Principle)

Archimedes' Principle states that any object submerged in a fluid experiences an upward force equal to the weight of the fluid it displaces. This upward force is a critical factor in understanding the net force acting on submerged bodies and how it affects their acceleration and depth of penetration in fluids.

Conservation of Mechanical Energy

This concept involves the transformation of gravitational potential energy into kinetic energy (and vice versa) when an object moves under the influence of gravity without energy losses. It allows us to determine speeds and heights by equating potential and kinetic energies, assuming that non-conservative forces (like friction) are negligible or absent.

Transcript

00:01 Hi there, so for this problem, for part a of this problem, we are asked with the information that we are given to find the speed at which the ball enters the liquid.

00:13 Okay, so for that we are going to let the ball be released at point one, enter the liquid at point two, attain a maximum death at point three, and pop through the surface on the way up at point.

00:31 So with that said, what we are going to use to calculate this speed is to use the energy conservation for the valve through the air.

00:40 So we will have that the sum of the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy plus the final potential energy.

00:52 Then we substitute the expressions in here for each energy.

00:59 So we know that the initial kinetic energy is equal to zero because the initial speed is zero.

01:05 Then we will have the mass types of acceleration due to gravity times the height at the point one.

01:11 This is the gravitational potential energy.

01:14 And then this is equal to the final gravitational potential energy in this case is zero.

01:20 So this equals to 1 divided by 2 times the mass times the speed at the point 2 to the square.

01:26 So we just need to simply solve for the speed.

01:30 We can cancel this with this one.

01:32 So the speed of the point two is the squared root of two times the acceleration to gravity times the height one.

01:38 And we use the values that we are given.

01:40 Two times the acceleration of gravity, that is 9 .8 meters per second square.

01:45 And the height one that we are given, that is 3 .3 in units of meters.

01:51 And we take the square root of all of this.

01:52 So using our calculator, we obtain a value of 8 .04 meters per second.

02:00 So that's a solution for part a of this problem.

02:03 Now for part b, we are asked about to evaluate the magnetis of the two porches that are inserted on the ball, as it moved through the liquid.

02:18 So that means the gravitational force and the buoyant force for this.

02:25 So the gravitational force is just the product between the mass and the acceleration due to gravity.

02:32 We're given the mass that is 2 .1 kilograms.

02:35 And the acceleration to gravity, again, is 9 .8 meters per second square.

02:40 So from this, we obtain a force that is 20 .6 neutons, but this is down, because this is the weight that always points down.

02:50 And now, for the buoyant force, we know that the buoyant force is defined as the mass of the fluid, this times the acceleration to gravity.

03:01 Now, we can write the mass of the fluid as its density.

03:06 The density of the fluid, this times the volume of the object times the acceleration due to gravity.

03:13 Then from this we will have the density of the fluid, this 4 divided by 3 times pi times the radius to the 3 times the acceleration due to gravity.

03:27 And now we just need to simply substitute all of the values in that expression.

03:30 So the density for the fluid we are given that is 1 ,000.

03:38 230 units of kilograms per cubic meter.

03:43 This times 4 times pi divided by 3.

03:47 And this times the radius that is 0 .09 meters and that to the 3.

03:55 And this times the acceleration due to gravity that is 9 .8 meters per second square...

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