A small mass m hangs at rest from a vertical rope of length ℓthat is fixed to the ceiling. A for

A small mass m hangs at rest from a vertical rope of length...

A small mass $m$ hangs at rest from a vertical rope of length $\ell$ that is fixed to the ceiling. A force $\vec { \mathbf { F } }$ then pushes on the mass, perpendicular to the taut rope at all times, until the rope is oriented at an angle $\theta = \theta _ { 0 }$ and the mass has been raised by a vertical distance $h$ (Fig. $34 ) .$ Assume the force's magnitude $F$ is adjusted so that the mass moves at constant speed along its curved trajectory. Show that the work done by $\vec { \mathbf { F } }$ during this process equals $m g h ,$ which is equivalent to the amount of work it takes to slowly lift a mass $m$ straight up by a height $h .$ [Hint: When the angle is increased by $d \theta$ (in radians), the mass moves along an arc length $d s = \ell d \theta . ]$

Key Concepts

Integration Along a Curved Path

When a force is applied along a curved path, the total work done is determined by integrating the infinitesimal contributions of work along each segment of the path. This requires relating small changes in the path's parameter (such as an angle in circular motion) to the actual displacement through differential elements like the arc length, ensuring that the work calculation accounts for the geometry of the motion.

Constant Speed Motion and Energy Conservation

In scenarios where an object moves at constant speed, there is no change in its kinetic energy, indicating that the net work done on the object is zero. However, when dealing with opposing forces (such as an applied force counteracting gravity), the work done by one force (here, raising the object) is exactly converted into an increase in potential energy. This balance underscores the principles of energy conservation in systems with balanced forces.

Work Done by a Force

Work is a measure of energy transferred by a force acting through a distance. It is calculated as the dot product of the force vector and the displacement vector, meaning only the component of the force in the direction of the displacement contributes. In many physical situations, such as lifting an object, the work done by the force directly results in energy changes in the system.

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object due to its position within a gravitational field. When an object is raised by a height h, its potential energy increases by an amount equal to mgh, where m is the mass and g is the acceleration due to gravity. This concept is central in understanding energy conservation when forces act over vertical displacements.

Transcript

00:01 7 .91.

00:02 So we have a mass on a rope that's attached to a ceiling or something, and we have a force that's always acting perpendicular to the rope that the mass is on, that lifts it up to some height, h.

00:17 And we want to show that the work that this force does is equal to just the work of lifting the mass, a height h.

00:27 And we're given a hint that when the angle is changed by v - theta, and radians, the arc length along which the mass moves, so along the sort of circle here, is l d theta.

00:51 This basically just follows from the definition of an angle, which is the arc length along a circle subtended by that angle divided by the radius of the circle.

01:02 So d theta is ds over l, or ds equals l d theta.

01:10 So we have this force.

01:12 We also have the weight of the little mass here.

01:20 And this is done without the mass accelerating.

01:29 We also have a force of tension.

01:32 And at any point along this path that it's taking, we can imagine we have an x and y directions.

01:44 The actual orientation of these changes as you increase the angle or decrease the angle.

01:53 For convenience, let's just think of them as x and y.

02:01 So the sum of the x forces, we have f acting in the positive x direction, and then we have a component of the x component of the weight that ends up acting in the negative x direction.

02:19 This is mg sine theta.

02:30 Because it's not accelerating, this has to be equal to zero.

02:40 So the force is mg sine theta.

02:43 So far so good.

02:47 Now the work that this force does will be the integral, well, let's not worry about the limits just yet, of f .d .l.

03:10 Or ds, whatever.

03:12 Let's call it ds, actually, because l already has a use of this problem as a constant...

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