Boxes are transported from one location to another in a...

Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of 0.50 $\mathrm{m} / \mathrm{s}$ At a certain location the conveyor belt moves for 2.0 $\mathrm{m}$ up an incline that makes an angle of $10^{\circ}$ with the horizontal, then for 2.0 $\mathrm{m}$ horizontally, and finally for 2.0 $\mathrm{m}$ down an incline that makes an angle of $10^{\circ}$ with the horizontal. Assume that a 2.0 $\mathrm{kg}$ box rides on the belt without slipping. At what rate is the force of the conveyor belt doing work on the box as the box moves (a) up the $10^{\circ}$ incline, (b) horizontally, and (c) down the $10^{\circ}$ incline?

Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of 0.50 $\mathrm{m} / \mathrm{s}$ At a certain location the conveyor belt moves for 2.0 $\mathrm{m}$ up an incline that makes an angle of $10^{\circ}$ with the horizontal,
then for 2.0 $\mathrm{m}$ horizontally, and finally for 2.0 $\mathrm{m}$ down an incline that makes an angle of $10^{\circ}$ with the horizontal. Assume that a 2.0 $\mathrm{kg}$
box rides on the belt without slipping. At what rate is the force of
the conveyor belt doing work on the box as the box moves (a) up
the $10^{\circ}$ incline, (b) horizontally, and (c) down the $10^{\circ}$ incline?

Key Concepts

Force Decomposition on an Inclined Plane

When dealing with an inclined plane, the gravitational force acting on an object is resolved into two components: one perpendicular to the plane and one parallel to it. The parallel component, given by mg sin(?), is the portion of the weight that must be overcome by an applied force to move the object upward or is assisted when moving downward.

Conservation of Energy/Work-Energy Principle

The work-energy principle states that the work done on an object results in a change in its energy, be it kinetic or potential. In scenarios of constant speed, any work done by an external force is used to change the gravitational potential energy for ascent or is counteracted by gravity during descent.

Work

Work is the energy transferred when a force is applied to move an object over a distance. It is calculated as the product of the component of the force in the direction of the displacement and the displacement, measured in Joules.

Power

Power is the rate at which work is done or energy is transferred, expressed as work done per unit time. In contexts where an object moves at constant speed, power can be computed as the product of the effective force in the direction of motion and the constant velocity.

Constant Velocity Motion

Constant velocity motion implies that there is no net acceleration, meaning that the forces acting on the object are in equilibrium. This allows the analysis to focus on the balance of forces, particularly when determining the force required to overcome gravitational components on an incline or maintain horizontal motion, thereby simplifying the calculation of power.

Transcript

00:01 All right, so in this problem, so we have a conveyor belt, but there are three processes.

00:09 The first one has an incline with the 10 degrees above the horizontal plane, and then this is a horizontal plan, and the last one is going down, and the face, and this angle is the same, 10 degrees.

00:31 So we already know the cargo is moving with constant velocity v equal to 0 .5 meters second and the mass of the cargo is 2 kilograms and so okay so we want to find we want to find out the the power of the conveyor belts in the three in the three processes so for the the first one for the going up for the going up phase the power will be equal to f times v, right? so f is the force inserted by this balance.

01:15 But we do not know it yet.

01:16 But the v is simply equal to 0 .5, so equal to 0 .5 times f.

01:22 So what will be the f in this case? so just look at this incline.

01:28 So this is the particle and this is the gravitational force.

01:33 You see that the reason why this this this cargo is stay here without moving down is because of the friction.

01:43 So the friction is basically moving up, it is pointing up, and the parallel to this incline.

01:52 So this is the force that does work to this cargo.

01:59 So the amount of this force is actually equal to 0 .5 times m g times co.

02:09 Sorry, times sine theta, where this data is 10 degrees.

02:18 That's because if you just do the force analysis, you can easily say that.

02:25 So this is the force parallel to the incline from the gravitational force, and this force is actually equal to the friction.

02:37 So the amount of this force is actually equal to the magnitude, of the magnitude of the gravity times sign this degree.

02:48 And this angle is actually equal to this angle...

Please give Ace some feedback