A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0^∘above the horizontal

step 1 This is Chapter 6, Section 1 Problem Number 11. We have a carton on an inclined plane, and it is

A 128.0-N carton is pulled up a frictionless baggage ramp...

Key Concepts

Work

Work is a measure of energy transfer that occurs when a force causes a displacement. It is calculated as the dot product of the force and displacement vectors, meaning that only the component of the force in the direction of the displacement contributes to the work done.

Dot Product

The dot product is a mathematical operation that multiplies the magnitudes of two vectors and the cosine of the angle between them. In the context of work, this operation determines how much of the force is effectively used in moving an object in the direction of its displacement.

Force Decomposition

Force decomposition involves breaking a force into components along chosen directions, often parallel and perpendicular to the direction of motion. This is especially important on inclined planes, where gravitational and applied forces must be separated into parts that do or do not contribute to motion along the incline.

Gravitational Work

Gravitational work refers to the work done by the gravity force as an object moves in a gravitational field. Since gravity is a constant force acting downward, its contribution to work depends on the vertical displacement and the angle between the displacement and the gravitational force vector.

Normal Force Work

The normal force is the force exerted by a surface perpendicular to an object in contact with it. In many scenarios, like motion along an inclined plane, the displacement of the object is perpendicular to the normal force, which means the normal force does no work because the cosine of 90° is zero.

Work-Energy Theorem

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. This theorem provides a direct connection between the forces acting on an object and its resulting motion, allowing calculation of speed changes without detailed analysis of the motion itself.

Force Orientation

Force orientation refers to the angle at which a force is applied relative to a given displacement. When a force is not aligned with the displacement, only its component along the path contributes to doing work, highlighting the need to use trigonometric projections when calculating the work done by forces applied at an angle.

Transcript

00:01 This is chapter 6, section 1 problem number 11.

00:04 We have a carton on an inclined plane, and it is being pulled up by a rope.

00:12 So this is our inclined plane.

00:14 It is at a 30 -degree angle, and it's being pulled up by a rope.

00:23 So let's call this force f sub r.

00:28 And let's show the other forces acting on this carton.

00:32 So the gravitational force is pointing down as you'll know.

00:37 And then the normal force is going up f sub n.

00:42 So this is the free bedded diagram.

00:46 So we're showing all the forces acting on the carton.

00:48 So one thing that is given to us in the problem is also that this f sub r is causing a displacement.

00:56 The direction of the displacement is actually up the inclined plane.

01:00 So let's note it, let's use the letter s to refer to the displacement vector.

01:07 If the initial position is something like this, then the final position of the carton would be out there so that this is the direction of displacement vector.

01:17 It's determining the direction of displacement vector is crucial in order to calculate the work because the general formula, let's remember work equals to the force, dot product, the displacement vector.

01:34 When we open this up, it's going to be f magnitude of the force, magnitude of the displacement vector, and the cosine of the angle between these two vectors, f and s.

01:49 So since we're asked in all the parts of this problem, we're asked to calculate the work done, basically this is of calculating, figuring out what theta is is going to be really important for us.

02:02 So in part a, we are to calculate the work done by the rope.

02:08 So then if you're calculating the work done by the rope, what we have is the force exerted by the rope times the magnitude of the displacement vector and the cosine of the angle between these two vectors.

02:24 So if you go back and look at a free battery diagram, the direction of the the displacement vector is up the inclination, and so is the direction of the force exerted by the rope.

02:38 So then, as you can see, they're parallel with respect to each other, which means the angle between these two vectors is actually zero, so then cosine zero would give us one.

02:55 Then we're calculating the work done by the rope.

02:59 All we need to do is actually plug in the magnitude of force exerted by the rope which is given to us as 72 newtons in the problem so 72 newtons times the magnitude of the displacement 5 .2 meters and cosine zero which is one so that equals to 374 .4 jules.

03:31 Now in part b we are asked to calculate the work done by gravity then that's going to be equal to the the magnitude of the gravitational force times the displacement, a magnitude of the displacement vector times the cosine of the angle between f sub g and s.

03:52 So going back to the free body diagram, this is the direction of the gravitational force.

04:00 And if we carry the displacement vector here, we can see the angle between these two vectors, f sub g and s, is 90 plus 30 degrees.

04:14 So the theta here is 90 plus 30 which is 120.

04:24 Then f sub g the magnitude is given to us as 128 newtons.

04:33 The magnitude of the displacement vector is 5 .2 and then we have cosine 120 degrees.

04:46 So if we do a calculation correctly, what we get is something negative due to cosine 120 being negative.

04:55 So 332 .8 joules would be the work done by gravity.

05:06 So in part c, we are to calculate the work done by the normal force.

05:14 Again, normal force times the magnitude of the displacement vector times the cosine of the angle between these two vectors.

05:27 If we look at the free by the diagram, this is the normal force, and this is the direction of the displacement vector.

05:34 As you can see, there is 90 degree angle between these two vectors...

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