A 5-kg collar A is at rest on top of, but not attached to, a...

Key Concepts

Kinetic Energy

Kinetic energy quantifies the energy that a mass possesses due to its velocity, expressed as 1/2 times the mass times the square of its speed. In systems where forces accelerate masses, understanding how much energy is manifested as kinetic energy is key to establishing energy balances.

Work Done by a Constant Force

When a constant force acts along the direction of displacement, the work done is given by the product of the force and the displacement. This concept connects the external input of energy into the system with the resulting changes in kinetic and potential energy, such as those observed when compressing a spring.

Hooke's Law and Elastic Potential Energy

Hooke's Law establishes that the force exerted by a spring is proportional to its displacement from equilibrium, with the spring constant determining the stiffness. The energy stored in a compressed or stretched spring is calculated as 1/2 times the spring constant times the displacement squared, making it essential for solving problems involving spring compression.

Work-Energy Principle

This principle states that the work done by external forces on a system is converted into changes in the kinetic and potential energy of that system. It is crucial in analyzing how the energy applied by a force leads to both the motion of a mass and the storage of energy in compressed or extended springs.

Transcript

0:00 Hi everyone.

00:01 What we have right now is a collar of mass 5 kilograms that is not attached to a spring k1, which has a spring constant 400 nm per meter, and there is a constant force of 150 newtons that is applied via a cable.

00:18 More as to find is the spring constant of the second spring, knowing that the velocity is 1 meter per second when the spring k1 is compressed 75 millimeters.

00:29 So what this looks like in a diagram is we have our top spring of k2, which is what we're trying to solve for.

00:41 We have a bar, and then we have a collar on this bar.

00:47 And this collar is attached to or is resting on top of spring k1, which we know.

00:57 And these are our distances.

00:58 We know that this is 450 millimeters, and the spring k -1 is compressed 75 millimeters, and at that point we know that the velocity of a, which i'm just going to call it right here, is one meter per second when it's being pulled on top of a cable that has a force of 150 nons.

01:22 And there is a pulley here, but we're going to assume that the mass of the pulley and the friction or negligible, so we're going to ignore those.

01:30 So what we're going to do is use the method of work energy and apply to caller a.

01:35 So we're simply going to say that kinetic energy at 1 plus the potential energy from 1 to 2 is going to equal the kinetic energy at 2.

01:44 We know that the caller is starting at rest.

01:47 So t1 is just going to be equal to 0.

01:50 So now we're just going to analyze position 2.

01:53 We know that in position 2, the upper spring is going to be compressed 75 millimeters.

02:01 Remember that is a given statement that we have.

02:04 And at that point, we know that the velocity is also 1 meter per second.

02:08 So we know that t2 is just going to be equal to 1 half mv2 squared.

02:13 Remember that we know the mass is 5 kilograms, you know the velocity, so we can just get this as 2 .5 joules.

02:21 So now we're going to look at it when the collar is raised from basically one to two, from position a to position b.

02:31 The force of the weight is going to be equal to negative mgh, where mass is 5 kilograms, g is 9 .1, 8 .9 .81 meters per second squared, and the height is 450 millimeters, because that's also given to us in the problem.

02:51 And what this is going to result in is a change in potential energy equal to negative 22 .025 22 .0725 joules.

03:05 So we know that in position 1, the force exerted by the lower spring is equal to the weight of the collar, and you know, we can just draw a free by a diagram to show that.

03:14 In position 1, we have the weight of acting down on the collar, and then we have the spring force of k1.

03:22 Pushing out and i'm going to call that ffs.

03:26 So we can rewrite this as saying that f1 or fs is going to be equal to the weight and that is equal to negative mass times gravity and that's going to be equal to negative 49.

03:44 49 .5 newtons.

03:48 Now as we know, now we know that when the collar moves up the distance x, the spring force is going to change and that's simply going to be equal to f s minus k x and that fs is what we saw for here that negative 49 .05 noons and it's going to be that until the collar separates at xf which is going to be f1 over k1 so if we solve for that we can say that xf is going to be equal to f1 over k1 which will be equal to we know f1 is 49 .05 newtons.

04:24 We know k1 is 400 ms for meter and this will translate into 122 .625 millimeter.

04:33 Now we can solve for the work exerted by the spring force by using this expression.

04:41 So we can see that u from 1 to 2 of spring 1 is equal to 0 x f1 minus k1 x d x d x.

04:52 And this integral.

04:54 So for plugging in the quantities that we know, actually we can just go ahead and solve this integral in terms of its variables.

05:01 So we can say that this is equal to f1 minus k1x squared over 2, evaluated from 0 to xf...

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