A horizontal block-spring system with the block on a...

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 51.4 J and a maximum displacement from equilibrium of 0.287 m. (a) What is the spring constant? 1248.042 N/m (b) What is the kinetic energy of the system at the equilibrium point? 51.4 J (c) If the maximum speed of the block is 3.45 m/s, what is its mass? 8.637 kg (d) What is the speed of the block when its displacement is 0.160 m? 8.20 m/s (e) Find the kinetic energy of the block at x = 0.160 m. J (f) Find the potential energy stored in the spring when x = 0.160 m. J (g) Suppose the same system is released from rest at x = 0.287 m on a rough surface so that it loses 12.6 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant? m Need Help? Read It

Transcript

0:00 All right, hello.

00:01 In this question, we're told that we have a spring block system.

00:04 And we're told that the total energy of the system is 51 .4 joules.

00:09 And our maximum amplitude, or maximum compression, or stretch distance is 0 .287 meters.

00:16 And we're asked, part a, what is the spring constant? well, we know that when we're fully stretched out, all of our energy is going to be equal to our spring potential energy.

00:25 And all of that is going to be 1 and 1 half times k delta x squared.

00:28 So in this case, k is going to be our total energy times 2 divided by our maximum distance here squared.

00:36 And we have all of those values.

00:37 So we can go ahead and plug them in.

00:39 And if we do that, we'll get a k value of 1 ,250 newtons per meter.

00:46 We're then asked, what is the kinetic energy at the equilibrium point? so at the equilibrium point, if we stretch the spring over to some delta x max over here, i guess this would technically be delta x max.

00:57 And then we let go.

01:00 When it goes back through this equilibrium point, it's going to be moving at its maximum speed because all of its energy is going to be equal to the kinetic energy.

01:09 All of its energy is going to be, there's no spring potential at all.

01:13 So it's just going to be kinetic energy in that case.

01:16 And so part b is, what is the kinetic energy of the system at its equilibrium point? well, at its equilibrium point, the energy of the system is all kinetic energy.

01:26 So its kinetic energy at that point is going to be 51 .4 joules.

01:31 Part c then says, what is the maximum speed? or if we know what the maximum speed is, so we know that v max is equal to 3 .45 meters per second, what is our mass going to be? well, we know at the equilibrium point, our kinetic energy is equal to 51 .4 joules.

01:51 And we know kinetic energy is 1 half mv squared.

01:55 So if we want to solve for m, we're going to have 2 times our kinetic energy divided by our speed squared.

02:00 We have all of those.

02:02 And so if we go ahead and put that in our calculator, we get a mass of 8 .6 kilograms.

02:08 We're then asked, what is the speed of the block when its displacement is 0 .6? so for this case, we have a delta x of 0 .16, sorry, not 6, 160 meters.

02:20 And we want to find what our speed is.

02:22 So we know that the total energy of our system is going to be spring potential plus the kinetic energy of the system.

02:29 In this case, we have some spring potential, 1 half k times delta x squared.

02:35 And we're going to have some speed, presumably, 1 half mv squared there.

02:39 We know what our total energy is, so we can go ahead and solve this for v.

02:43 And we're going to get an expression that looks like this...

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