Could somebody please help, I don't even know where to start with this. Please explain your answers to me, it would help me a lot

Question

Two masses (m1 = 2.0 kg, m2 = 4.0 kg) are each initially 2.0 m above the table, and the massless

frictionless pulley is 6.0 m above the table.

(a) What’s the acceleration of the mass 2?

(b) What’s the speed of the mass 2 at the moment when it hits the table?

(c) What’s the maximum height the mass 1 can move up from the table?

Answer 1

This is an interesting problem, and one you would do well to master. Setups like this are called 'Atwood Machines' (you can look these up).

First, the acceleration of mass 2. For that, you need to find the net force on the system.

The weight of m₁ creates a force F₁=m₁g, that is pulling m₁ down, working to rotate the pully counterclockwise. Meanwhile, the weight of m₂ creates a force of F₂=m₂g that is pulling m₂ down, and working to rotate the pully clockwise. Because m₂ > m₁ the net force will be causing m₂ to fall and rotating the pully clockwise, and the net force on the system will be F_net = (m₂-m₁)g.

To find the acceleration, then, you just need to use F = ma, to find the acceleration of the system. Since you're accelerating both masses, and the pully is massless and frictionless (letting you ignore it), then m_total = m₁ + m₂. That gives

a_net= (m₂-m₁)g/(m₁ + m₂).

I leave the math to you.

To find the speed of m₂ when it hits the table there are actually two approaches you can take. First, you can use kinematics, and (v_f )² = (v_o)² + 2aΔx. Knowing a from above, that v_o= 0 and that the distance Δx is 2 meters, you can find v_f. You do the math...

Or, you can use the change in energy of the system. You have to add energy to m₁ to raise it (m₁gΔx), and energy is released from m₂ as it falls (m₂gΔx). So,

E_net = (m₂-m₁)gΔx.

That energy has become the kinetic energy of the system, so

E_net = (m₂-m₁)gΔx = (1/2)(m₁+m₂)(v_f)²

Again, I leave the math to you. Assuming all my algebra is correct, you can compare the two methods, to ensure they yield the same result.

The last question is how high can m₁ go. one simple answer is 2 m, as m₂ will be blocked by the table. However, it could momentarily rise higher, as it will have v_f when m₂ hits the table. That means kinetic energy, that will allow it to rise a bit higher, by some amount h, such that m₁gh = (1/2)m₁(v_f)².

TBH, I've not yet checked my work by 'doing the math' but the ideas should be correct. Message me if you need more information or want to go over this.

Answer 2

This problem is dealing with the workings of Free-Body Diagrams (FBDs) and the Kinematic Equations of Motion.

The key concept with FBDs is that Newton's Second Law of Motion applies:

∑F_sys = F_{net, sys} = m_sys * a_sys

The summation of all forces on the system, commonly rewritten as the total net force on the system, will cause the mass of the system to accelerate at the value of consequential total system acceleration.

You will almost always come out with 2 of the 3 variables in that equation (m and F find a, m and a find F, or a and F find m)

A convenient feature of this equation is that it can be applied in each of the principle directions, x, y, or z. So let's make up and down the y-direction and work with the following:

F_{y-net, sys} = m_sys * a_{y sys}

As you handle and apply different systems, the equation will appear to take on different forms.

Because you are asked to find the acceleration on mass 2, let's treat mass 2 as our current system and observe the forces acting on it.

Because this is on Earth, it will always have weight (m₂g) that will be pulling it down and the tension in the pulley rope trying to pull mass 2 up. Give this tension force some value T.

For the mass 2 system:

T - m₂g = m₂a₂

Likewise, mass 1 is also being pulled down by its own weight (m₁g) and pulling pulled up by the same tension force, T.

For the mass 1 system:

T - m₁g = m₁a₁

Now, this is 2 equations and 3 unknowns (a₁, a₂ and T). A third equation is needed to obtain numerical values for all 3.

Something that I was taught about these problems is to actually consider the rope (but not super technically). What you must consider is that the rope length will never change (he said optimistically) under the low forces presented in this problem. In other words, the length of the rope is constant.

To describe the length of the rope, you must make it in reference to the heights of each box. Let's call these heights y₁ and y₂ and measure these heights from the table. The pulley itself is 6 meters above the table. To get the left portion of the pulley rope for mass 1:

6 m = Y_L + y₁ Y_L = 6 - y₁

For the right side of the rope for mass 2:

6 m = Y_R + y₂ Y_R = 6 - y₂

And there is some rope around the pulley. It is the arc length across a half of a circle (basically half of the circumference). It is over 180° or π radians at the radius of the pulley (let's call it R). So, the arc length will look like:

L_P = πR

So, the length of the rope is the length of the left side of rope plus the right side of the rope plus the length of rope around the pulley. Mathematically stated:

L = Y_L + Y_R + L_P

Replace with what we have and get:

L = 6 - y₁ + 6 - y₂ + πR L = 12 + πR - y₁ - y₂

Now, if the length of the rope is constant, then if we take the derivative in respect to time on both sides of the equation, the derivative of ANY of the constant terms will go to zero and any of the variable terms will go to velocities.

L, R, π, and 12 are constant so:

0 = 0 + 0 - v₁ - v₂ 0 = - v₁ - v₂

Take another time derivative of both sides and any velocities turn into accelerations:

0 = - a₁ - a₂ a₁ = - a₂

What this says is that, as a consequence of the rope length being constant, whatever the acceleration of 1 block is, the other block will have the same magnitude of acceleration but in the exact opposite direction. It is also the 3rd equation for the 3 unknowns.

Here are the 3 equations to be working with:

T - m₂g = m₂a₂

T - m₁g = m₁a₁

a₁ = - a₂

Take the second equation of the 3 and rearrange like so:

T = m₁g + m₁a₁

Now, substitute a₁ for -a₂ in this new equation like so:

T = m₁g - m₁a₂

Next, substitute this new equation for T into equation 1 like so:

m₁g - m₁a₂ - m₂g = m₂a₂ (m₁ - m₂) * g = (m₁ + m₂) * a₂

a₂ = [ (m₁ - m₂) / (m₁ + m₂) ] * g

This has explicitly found the acceleration on block 2 in terms of m₁, m₂, and g

What should be noticed is that it was never assumed we would get a 100% positive-sign result. In saying that F_net = ma. It is initially guessing that the acceleration on mass 2 would be up (that is why there is no negative sign present on our initial F_net = ma equation), but it is wise to let the algebra really tell which direction it will end up being. Since mass 2 is greater than mass 1, mass 1 minus mass 2 will be a negative result. This is to say that mass 2 will have a negative acceleration upward. Another way to say that is that it will accelerate downward at the magnitude of the acceleration discovered through the equation above. Inputting the known values looks like:

a₂ = [ ( 2 kg - 4 kg ) / ( 2 kg + 4 kg ) ] * ( 9.81 m/s² ) = - 1 / 3 * 9.81 m/s² a₂ = - 3.27 m/s²

Next, there is a kinematic equation that exists that if you know the system's:

1. Acceleration
2. Displacement
3. Initial velocity

You can find the final velocity using the following equation:

v_f² = v_i² + 2 * a * Δd

Since the blocks start from rest, the equation simplifies to the following:

v_f² = 2 * a * Δd

Since block 2 is heavier and already has the negative acceleration, it will FALL the 2 meter height and NOT rise by that much height, so Δd = - 2 m.

v_f = √( 2 * (-3.27 m/s²) * ( - 2 m ) ) = √( 13.08 m²/s² ) v_f = ±3.62 m/s

To rectify the plus or minus, you must consider the situation at hand; is the block rising up into (or away) from the table or is it falling down into (or below) the table? It is falling down, so the correct answer would be: v_f = - 3.62 m/s

Intuitively, since the two masses are connected by the same rope, if one falls by 2 m then by consequence, the other must rise by that same 2 m to conserve the constant length of the rope. To prove this, we can also check through the kinematic equations. In the time that mass 2 fell 2 meters, is the same time that mass 1 rose some unknown Y meters.

Δd = v_i * Δt + 1 / 2 * a₂ * Δt² Δd = 1 / 2 * a₂ * Δt² Δt = √( 2 * Δd / a₂ )

Y = v_i * Δt + 1 / 2 * a₁ * Δt² Not to forget that: a₁ = - a₂ Y = 1 / 2 * ( - a2 ) * ( √( 2 * Δd / a₂ ) )²

Y = - Δd = -1 * ( - 2 m ) Y = + 2 m

I hope this helps!