2. A spring with a force constant of 500 N/m is resting on a...

2. A spring with a force constant of 500 N/m is resting on a frictionless surface and mounted horizontally on a wall. A 2.5 kg box is pushed against the spring, compressing it 10.0 cm from equilibrium. When released, the spring pushes the box across the floor. a. How much force must be exerted on the spring to compress it 10.0 cm? b. How much work is done on the spring to compress it 10.0 cm? c. How much elastic potential energy is stored in the spring while it is compressed? d. Once released, what maximum speed will the box attain?

Transcript

00:01 All right, so we have a horizontal spring with a mass compressed against it of 55 grams, so 0 .055 kilograms.

00:11 And it's compressed against the spring a distance of 11 centimeters, so 0 .11 meters.

00:17 And then lastly, the spring has a spring constant of 100 newtons per meter.

00:22 And so we want to know when it's released from this compression.

00:26 First off, what is the speed of the box at the instant it is passing the point of being compressed 7 centimeters? so it's moved 4 centimeters away from the point of maximum compression.

00:37 So what we'll have is the change in kinetic energy plus the change in potential energy is equal to zero.

00:43 So the initial kinetic energy of this system would be zero.

00:47 And the final kinetic energy, we'll just call this 1⁄2mv squared, where v is the velocity we want to find.

00:53 The initial potential energy is going to be like one -half k, we'll call this x -1 squared.

00:59 It's the initial distance that's compressed.

01:01 The final kinetic energy is one -half k x -2 squared, where it's like the second distance, the seven centimeters that we're asked to look at.

01:10 So if we look at this, what we'll have is one -half mv squared equals one -half k x -1 -squared minus x2 squared.

01:21 And so v is going to be the square root of, let's write it this way, k over m times x1 squared minus x2 squared.

01:32 So let's plug in our numbers.

01:34 We have 100 newtons per meter for our spring constant, and our mass was 0 .055 kilograms.

01:42 And then x1 was 0 .11 meters.

01:47 So actually, let me write this sort of, in brackets.

01:51 And then x2 is 0 .07 meters squared.

01:56 So this is what we're looking at.

01:59 And so if we got 0 .11 squared minus 0 .07 squared times 100 newtons per meter divided by 0 .055 and then we take the square root should get something like 3 .62 meters per second.

02:14 So that's our velocity at that point.

02:16 Part b says, what is the maximum speed and where does it occur? so the maximum speed occurs at the equilibrium point.

02:25 And so we can actually calculate it because it's just going to be, you know, the answer to the previous problem if we just set x2 equal to zero.

02:33 So it'll be like the square root of k over m times x1...

Question

Please give Ace some feedback