How Do Sliders A and B Behave Under Horizontal Force in Physics?

[dietwater]

1. Homework Statement

Each of the sliders A and B has a mass of 2 kg and moves with negligible friction in its
respective guide, with y being in the vertical direction (see Figure 3). A 20 N horizontal force
is applied to the midpoint of the connecting link of negligible mass, and the assembly is
released from rest with θ = 0°. Determine the velocity vA with which slider A strikes the
horizontal guide when θ = 90°.
[vA = 3.44 m/s]

2. Homework Equations

1/2 mv^2

F = ma

Wp = mgh

SUVAT


3. The Attempt at a Solution

When at 0 degrees

W=0J

At 90

F=20N

W = 20xd = 8J

Work from cart A = 0.5mv^2

Therefore 16 = mv^2

v = 2 rt2

Or...

do i need to add the energy from 20n force and from cart b...

0.5mv^2 (b) + 8J = 0.5mv^2 (A)

with F = ma, 20/10 a = 10 therefore v (b) = 2 rt 2

sub this into above eq.

8 + 8 = 0.5mv^2

v = 4

Help!

Iv been goin round in circles, clearly I am wrong lol can someone explain how i could work this out please

 

[Jhero]

?

Hi there! I can help you with your question. Firstly, let's define some variables to make things easier. Let's call the mass of slider A m_A, the mass of slider B m_B, and the force applied F.

Now, let's look at the system at 0 degrees. At this point, the force F is not doing any work since it is perpendicular to the displacement of the sliders. Therefore, the total energy of the system at this point is just the potential energy due to gravity, which is equal to m_Agh, where h is the height of the sliders.

Next, let's look at the system at 90 degrees. At this point, the force F is doing work on the system, causing both sliders to move. The work done by F is equal to Fd, where d is the displacement of the sliders. Since the force is applied to the midpoint of the connecting link, the displacement of both sliders is equal to d/2. Therefore, the work done by F is equal to Fd/2.

Now, let's use the work-energy theorem to relate the work done by the force to the change in kinetic energy of the system. The work-energy theorem states that the work done by a net force on an object is equal to the change in kinetic energy of that object. In this case, the net force is just the force F, and the object is the entire system of sliders A and B. So, we can write:

Fd/2 = 0.5(m_Av_A^2 + m_Bv_B^2)

where v_A and v_B are the velocities of sliders A and B, respectively.

Now, let's use the fact that the sliders have the same mass (2 kg) and that they start at rest, to simplify this equation. We can rewrite it as:

Fd/2 = 2v_A^2

Rearranging for v_A, we get:

v_A = √(Fd/4)

Now, we just need to plug in the known values. We know that F = 20 N and d = 2 m, so we get:

v_A = √(20*2/4) = √10 ≈ 3.16 m/s

So, the velocity of slider A when θ = 90° is approximately 3.16 m/s. This