A 40- -lb block B is suspended from a 6 -ft cord attached to a 60 -lb cart A, which may roll fre

A 40- -lb block B is suspended from a 6 -ft cord attached to...

Key Concepts

Conservation of Mechanical Energy

In many physics problems involving motion without non?conservative forces like friction, the total mechanical energy (the sum of kinetic and potential energy) remains constant. This principle allows one to relate changes in gravitational potential energy to the kinetic energies of the moving bodies, making it possible to solve for their speeds at various positions in the system.

Constraint Relationships in Coupled Systems

When bodies are connected by inextensible cords or rigid rods, their motions are not independent. The geometrical constraints imposed by the connection restrict the relative movements of the objects, which is expressed mathematically through relationships between their displacements, velocities, and accelerations. This is crucial in systems where one body’s movement affects the motion of another, such as a pendulum attached to a sliding cart.

Conservation of Linear Momentum

In a system where there are no external forces acting in a particular direction, the total linear momentum in that direction remains constant. In problems where some forces, such as weight, act vertically while the horizontal motion is free from external influences, this conservation law can be used to relate the horizontal speeds of different parts of the system.

Transcript

00:01 So in this problem, we are given the initial starting point of block b and the cart a.

00:07 And we wish to calculate the speed of the cart a and the block b at this position.

00:15 So we'll draw a diagram.

00:17 So cart a is moving horizontally to the right.

00:23 We'll choose this as positive with momentum m -a -v -a.

00:28 So it has velocity v -a.

00:31 Attached to the cart is block b.

00:38 So we're interested in the point at which block b passes directly under cart a.

00:43 And it's moving to the right with momentum mbvb.

00:49 So it has velocity at this point vb.

00:52 So the first thing we'll do is use the conservation of linear momentum.

01:02 We know that the momentum when it is released of the system is equal to the momentum as it passes directly.

01:10 Under the cart as the block passes directly under the cart.

01:14 So since the block in the cart i initially at rest, the initial linear momentum l -0 of the system is zero.

01:24 And as b passes under a, the linear momentum of the system is equal to that due to the cart m -a -v -a plus the momentum due to the block m -b -b.

01:43 So if linear momentum is conserved, this means m a va, va plus mb vb must equal to 0.

02:07 So from here, let's rewrite this expression and make va the subject of the formula.

02:14 So va is minus mb over m .a times vb.

02:25 And we'll call this equation 1 so we'll use this again later now let's look at the diagram and consider some of the trigonometry so initially we have cart a and that is the vertical the line at which the weight is acting and then we have the cord or the cable of length l and the block b so this is initially when the system is addressed this angle will call theta.

03:17 So the vertical distance, mark this in a different color in red, so this vertical height is l cost theta using some trigonometry.

03:43 And afterwards, we have block a moving to the right with velocity va, and the block a vertical distance directly below the, the cart a moving also to the right with velocity vb so it's a distance l from the cart so now let's apply the conservation of energy so initially the energy kinetic energy t -not is zero the system is at rest the potential energy of the block v0 is equal to m g h so the height of the block above its position directly below the cart a.

04:51 So this is the mass of the block times acceleration due to gravity tens h and h is this in this case is l into 1 minus cost theta.

05:06 So l minus al cost theta gives us this height above the rest position...

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