(II) Two masses are connected by a string as shown in Fig. 34 . Mass mΛ=4.0 kg rests on a fricti

step 1 to approach this problem with free body diagrams. So from these, this is the free body diagram f

(II) Two masses are connected by a string as shown in Fig....

(II) Two masses are connected by a string as shown in Fig. $34 .$ Mass $m_{\Lambda}=4.0 \mathrm{kg}$ rests on a frictionless inclined plane, while $m_{\mathrm{B}}=5.0 \mathrm{kg}$ is initially held at a height of $h=0.75 \mathrm{m}$ above the floor. $(a)$ If $m_{\mathrm{B}}$ is allowed to fall, what will be the resulting acceleration of the masses? $(b)$ If the masses were initially at rest, use the kinematic equations to find their velocity just before $m_{B}$ hits the floor. (c) Use conservation of energy to find the velocity of the masses just before $m_{B}$ hits the floor. You should get the same answer as in part $(b) .$

(II) Two masses are connected by a string as shown in
Fig. $34 .$ Mass $m_{\Lambda}=4.0 \mathrm{kg}$ rests on a frictionless inclined
plane, while $m_{\mathrm{B}}=5.0 \mathrm{kg}$ is initially held at a height of
$h=0.75 \mathrm{m}$ above the floor. $(a)$ If $m_{\mathrm{B}}$ is allowed to fall, what
will be the resulting acceleration of the masses? $(b)$ If the
masses were initially at rest, use the kinematic equations
to find their velocity just before $m_{B}$ hits the floor. (c) Use
conservation of energy to find the velocity of the masses just
before $m_{B}$ hits the floor. You should get the same answer as
in part $(b) .$

Key Concepts

Conservation of Mechanical Energy

The principle of conservation of mechanical energy states that in the absence of non-conservative forces (like friction), the total mechanical energy (the sum of kinetic and potential energies) remains constant. This concept is used to equate the loss in potential energy to the gain in kinetic energy, enabling the determination of speeds in systems where objects change height or configuration.

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. These equations relate displacement, initial velocity, final velocity, acceleration, and time, allowing us to determine the velocity or displacement of a system after a given time interval, especially when the acceleration is known from force analysis.

Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This is fundamental for analyzing systems with multiple objects, as it allows us to write equations for each mass and solve for unknown variables like acceleration or tension in connecting strings by balancing the forces along the direction of motion.

Free Body Diagrams

Free body diagrams are graphical illustrations used to visualize the forces acting upon a single object. By isolating an object and representing all the forces (such as gravity, normal force, and tension), free body diagrams help in setting up the equations based on Newton's laws, thus simplifying the process of analyzing complex systems like connected masses on different surfaces.

Transcript

00:01 To approach this problem with free body diagrams.

00:04 So from these, this is the free body diagram for block a and this is for block b.

00:10 And they're connected by a pulley, of course, so the tension force is the same.

00:14 So you resolve forces in the x direction for block a.

00:22 And what you get is that f of t, the tension force, minus mag sine theta, the x component of gravity is equal to, oops, not zero, is equal to ma times aa, that acceleration of a.

00:46 And right, and so that's the first equation.

00:50 Then you resolve forces in the y direction for block a, and what you get is fn, normal force is just ma times g cosine.

00:59 Uh, theta vertical component of gravitational force.

01:04 And then you resolve forces in the y direction for block b.

01:07 You have that, uh, well, so let's consider, so in this case, we'll consider, uh, the downward direction is positive.

01:15 So you have negative ft plus mbg is equal to mb times, uh, an acceleration of b.

01:23 So ft is actually mb times g minus a, b, b.

01:29 Okay? and so we have these three equations and so yeah let's call this equation one, this equation two and this equation three.

01:42 So in equation in equation one, therefore we can plug in three and so we have m a times a by the way a acceleration is due to the rope and so a a a is equal to ab is equal to a.

02:02 So m .a times a, which is m .a times a, is ft, which is mb times g minus a of b, plus minus m -a -g sine theta from the equation.

02:19 So a times m -a plus m -b, pulling a terms to the left -hand side, is equal to g times m -b minus m -a sine theta and so a is equal to g times m b minus m a sine theta over m a plus m b so we get 9 .8 meters per second squared times 5 kilograms minus 4 kilograms sine 32 over 5 plus 4 kilograms.

03:09 And so what you get out of this is acceleration is 3 .1 meters per second square.

03:18 Okay, all right, so that was part a...

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