a-horizontal-spring-is-attached-to-a-5-kg-block-when-the-block-is-pulled-5-mathrmcm-the-right-the-re

Question

A horizontal spring is attached to a 5 kg block. When the block is
pulled 5 $\mathrm{cm}$ the right, the restoring force has a magnitude of 6 $\mathrm{N}$ . What is the frequency of the spring?

Salamat Ali · Accepted Answer

the frequency off this spring can be found by f is equal to by finding spring constant. First Medusa, which can be found by using a hoax law that is a physical. Two cakes force we have is a six Newton. So the extension of spring we have this is important. Zero five meter and dividing these numbers gives us the care constant value, which is a 1 20 newton new temp. Our meter you do. And the frequency we can write for spring is one over to buy our square root off. He&#x27;s going to kill him, huh? We just found the cave to one substantive alien for a cane. This equation writing one or two by okay, is the 1 20 How do I did on my mosque you were five kg then simplifying this expression uses. Is there a 0.78 Howard&#x27;s frequency off the scream? So option sees the right answer

Averell Hause · Answer

we know that the oscillation frequency ah or the angular frequency would be cling to two pi times the frequency F. This would be equaling the square root of the spring constant divided by the mass. And now we know that the weight of the block and G must be equal to spring. Constant times the change in length in order for the system to be in translational equilibrium in the Y direction and so on the other side. This is due to Hook&#x27;s law and so the gravitational force would be equal to the force Is the force of the spring due to Hook&#x27;s law and so finding despairing, constant K This would be equaling the gravitational force divided by the change in the length and so the frequency F would be equal in 21 over two pi multiplied by the square root of the spring constant K divided by the Mass M. This would be equaling than 2 1/2 pi multiplied by the square root of M G and divided by Delta L divided by m the mass. Of course, the mass cancel out and this is equaling. Then 21 over two pi multiplied by the square boot of G over doubt out. And so we find that then the frequency is equal. In 1/2 pi multiplied by the square root of 9.80 meters per second squared. This will be divided by 0.20 meters. Let me find that the frequency is equaling. 23.5 hertz. That is the end of the solution. Thank you for watching.

Surjit Tewari · Answer

question number 88. There is an incline, and the object is on them. Plan the spring Forces acting off the incline. This is gay exports and wait component perpendicular to the in plain is and g cost Rita and this component off the waiters. M g sign Rita, this is the normal force. And so if the block is not sliding, then we can write m g Signed Rita is equal to the Is Spring Force. Yes. For the block tonight, slide mass is 30 k g value of G s 9.8 and in those sign inclination angle, 60 degree equals Okay into the stretch off spring, which is 0.5 meters. From this. We will get the spring constant cake will, too five Seattle nine to Newton, but made that

Zachary Warner · Answer

for this question, we&#x27;re told that a spring is hung from the ceiling. The mass of a block attached to it emits 0.45 kilograms. It was released from rest and the block drop 0.15 meters before momentarily moving back upwards. And it wants to know the spring concert of the spring. So the masses 0.45 But the displacement here is 0.15 um, divided by two. So, uh, this is because the the masses going to also Lee that&#x27;s the total stretch displacement of the spring. So the masses going toe Ossa late back and forth between half that. Okay, so at this moment, when it comes to rest before it starts going back, back up the force from gravity is equal to the force acting on the spring f sub s. So we just replace these with our conditions. The force of gravity is mass times acceleration of gravity. G. The force of the spring is K times X toasted toe. Find screen constantly. Just simply rearrange this equation and salt for K, we have K is equal to mass times gravity divided by X. Therefore, the spring constant K is equal to 58.8, and the units here would be willing to redraw that eight would be Newtons per meter so he can, Box said. It is our solution to a part. B wants to know the angular frequency of this oscillation. So the angular frequency of the oscillation Omega is by definition equal to the square root of K over in for a simple harmonic oscillating spree. Plugging in those values, we find that this is equal to 11.4 radiance per second tweaking box that is your solution for party.

Averell Hause · Answer

here for part A. We know that the frequency F is equaling 1/2 pi multiplied by the square root of the spring Constant K divided by the Mass M. This is equally 1/2 pi multiplied by the square root of 2000 Newtons per meter. This will be divided by 5.0 kilograms and we find that then the frequency F is approximately 3.18 hurts For part B. We can find the amplitude from energy conservation. This would be equal in the square root of the initial position squared, plus the initial velocity divided by omega quantity squared. And so this is equaling square root of 0.500 meat. Rather, my family&#x27;s 0.50 meters quantity squared plus are 1.0 meters per second, divided by two pi multiplied by 3.183 hurts again. We&#x27;re round at the very end and this terms where stand the square root and we find that then the amplitude A is equaling 2.0 707 meters. This is essentially 7.7 centimeters, and then for part c, we can see that the total mechanical energy would be equaling 2 1/2 times this spring, constant multiplied by the amplitude squared. This would be 1/2 multiplied by 2000 Newtons per meter multiplied by point 0707 meters. Quantity squared. And we find that then the total mechanical energy is equaling, then approximately 5.0 jewels. That is the end of the solution. Thank you for watching.

Averell Hause · Answer

we&#x27;re gonna applying Newton&#x27;s second law, The stretching of the spring, several one multiplied by the spring. Constant K equaling the gravitational force MGI. And we can say that delta else of one plus Delta else of to multiply by the spring Constant K equals two mg. So the delta else of two equaling 5.0 centimeters. We could say that then delta else of one plus Delta else of two multiplied by K minus that delta else of one multiplied by k equals two m g minus MGI. And so if then find that 0.5 meters multiplied by K equals M g so que over m equals than 9.80 meters per second squared, divided by 0.5 meters. This is giving us 196 seconds raised to the negative second power with both blocks attached. Giving the total mass of two times M omega would be equal. The swimmer What&#x27;s the square root? Divided by swearing of the spring, constant divided by two times the mass, and this is equaling. Then the square root of 1/2 multiplied by 196 seconds. Raised to the negative second, and this is giving us than 9.90 radiance her second. Then we find the frequency. This would be equaling the angular frequency divided by two pi or 9.90 radiance for second, divided by two pi. Giving us 1.58 hurts. That is the end of the solution. Thank you for watching.

Problem

$$a=18 t+15$$ Jane made an initial deposit to a …

Problem 34 Medium Difficulty

A horizontal spring is attached to a 5 kg block. When the block is
pulled 5 $\mathrm{cm}$ the right, the restoring force has a magnitude of 6 $\mathrm{N}$ . What is the frequency of the spring?

Answer

Related Courses

Cracking The AP Physics 1 Exam

Related Topics

Discussion

Top Educators

Recommended Videos

Watch More Solved Questions in Chapter

Video Transcript

Princeton Review

Cracking The AP Physics 1 Exam

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Cracking The AP Physics 1 Exam

Princeton ReviewCracking The AP Physics 1 Exam

Princeton Review

Cracking The AP Physics 1 Exam