A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a fricti

step 1 Hello my name is Debe. In this video we'll cover work in kinetic energy. So for this problem we

A small block with a mass of 0.0600 kg is attached to a...

A small block with a mass of $0.0600 \mathrm{~kg}$ is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. $\mathrm{P} 6.71$ ). The block is originally revolving at a distance of $0.40 \mathrm{~m}$ from the hole with a speed of $0.70 \mathrm{~m} / \mathrm{s}$ The cord is then pulled from below, shortening the radius of the circle in which the block revolves to $0.10 \mathrm{~m}$. At this new distance, the speed of the block is $2.80 \mathrm{~m} / \mathrm{s}$. (a) What is the tension in the cord in the original situation, when the block has speed $v=0.70 \mathrm{~m} / \mathrm{s} ?$ (b) What is the tension in the cord in the final situation. when the block has speed $v=2.80 \mathrm{~m} / \mathrm{s} ?$ (c) How much work was done by the person who pulled on the cord?

Key Concepts

Work and Energy

Work in mechanical systems is related to the transfer of energy when a force causes displacement. In circular motion scenarios where the radius changes, work is done by forces (such as a person pulling a cord) that alter the kinetic energy of the system. The work done can be calculated using the work-energy theorem by finding the change in kinetic energy of the object.

Conservation of Angular Momentum

Angular momentum conservation is a key principle in circular motion problems, particularly when no external torque acts on the system. When an object moves in a circular path and the radius of the circle changes, the angular momentum (given by mvr) remains constant, which implies a change in speed as the radius is varied. This principle helps explain the relationship between radius reduction and speed increase.

Tension as a Source of Centripetal Force

In systems where an object is attached to a string or cord and is undergoing circular motion, the tension in the cord can provide the required centripetal force to maintain the motion. By equating the tension to m(v²/r), where m is the mass of the object, one can determine the forces acting on the object in different scenarios, such as varying radii and speeds.

Centripetal Force

This concept refers to the net force required to keep an object moving in a circular path at a constant speed. It is directed toward the center of the circle and is calculated using the formula a? = v²/r, where v is the speed of the object and r is the radius of the circular path. Understanding centripetal force is essential in analyzing situations involving circular motion.

Transcript

00:01 Hello my name is debe.

00:01 In this video we'll cover work in kinetic energy.

00:05 So for this problem we have a mass of 0 .06 kilograms attached to a strength and this mass is rotating within a circle in the original situation with a radius of 0 .40 meters and at this point it has a tangential velocity of 0 .70 meters per second and now in the final situation when you pull on the strength the radius becomes smaller which is 0 .10 meters but the tangential tangential velocity of the mass it increases to 2 .80 meters per second okay so we want to know what is the tension on on the strength for each of those two situations so first we want we want to know the radio acceleration so let's recall that the radio acceleration as we're doing one let's just call it one it's going to be big one a square divided by r1 and again r1 is just the distance of the strength from the hole and d2 will be the one for the final situation okay so we could just plug in our values for b1 and r1 so we see that the radio acceleration for the first situation it's 1 .2 to 5 meters per second square okay so now for the first situation we could apply norton second law of motion so f -net, the net force on the mass, it's equal to the mass, that's its radio acceleration, which we just found.

02:08 So the only force acting on the mass is the tension of the strength.

02:13 So the tension for the first situation is going to be the mass, which is 0 .0600 kilograms.

02:22 That's the radio acceleration we just found.

02:37 So we see that the tension on the strength for the original situation, if you run the rate, to two significant figures, it's 0 .074 newtons.

02:58 For part b, we want to note the tension for the final situation after you pull on the strength.

03:05 So we're going to find the radio acceleration also...

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