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(III) $\mathrm{A} 2.0$ -kg block slides along a horizontal surface with acoefficient of kinetic friction $\mu_{\mathrm{k}}=0.30 .$ The block has aspeed $v=1.3 \mathrm{m} / \mathrm{s}$ when it strikes a massless spring head-on.(a) If the spring has force constant $k=120 \mathrm{N} / \mathrm{m},$ how faris the spring compressed? (b) What minimum value of the coefficient of static friction, $\mu_{\mathrm{S}},$ will assure that the spring remains compressed at the maximum compressed position? (c) If $\mu_{\mathrm{s}}$ is less than this, what is the speed of the block when it detaches from the decompressing spring? [Hint: Detach-ment occurs when the spring reaches its natural length $(x=0) :$ explain why 1

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A) 0.13 $\mathrm{m}$B) 0.77C) 0.5$m / s$

Physics 101 Mechanics

Chapter 8

Conservation of Energy

Work

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Ali A.

October 5, 2021

f u bitches, cant give the answers for free

Rutgers, The State University of New Jersey

Hope College

University of Winnipeg

Lectures

04:05

In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

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So in this problem, we have your block off Mass two kilogram is at this point the blockers off Master Kilogram, moving with the velocity off V nut which is given by I just 1.3 meter for a second and just strikes a spring which is kept here with a force constant off 1 20 neutering sperm eater. And the surface has got the kinetic friction coefficient off 13 It stays and it strikes a massless spring. Really velocity of important see meters per second. So what's gonna happen is that now this block is going to compress the spring so the spring is going to get compressed and it's going to come to this location. So our dislocation, you know, we have X not Richard's equals to zero. And at this location we have X f for the blow crack your travels a distance of except towards the right and comes to a stop, and the North is equal to 1.3 and then the block comes to rest. VF is equals to zero. Now we'll write the equation of conservation of energy for this non conservative four situation which says that the change off kinetic energy, plus the change of elastic potential energy, his equals to the work done by non conservative force. The vote done by non conservative forces. Actually the friction force, which is minus four. So friction times l the distance Trevor on the ground. So if he really in this equation Oh, yeah, there is one more thing is that this negative sinus because the force of friction is opposite to the displacement. So the block is moving in this direction and the force of friction is acting in this direction trying to stop you there for the work done. It's negative. So feel rather equation. We'll get Delta Cape less Delta U less force of friction times l. That equals to zero, and we'll use this equation here. So now I'll write the equation half Mm. The F squired Miners v North Squared, which is Delta K Delta. U is equals to have gay except squared minus X snorts, Quiet. You know, X. Nor does zero. And we know that re f zero bless force of friction is mg mu k x f minus x Not so the term wishful. Cancel the ex not, And then we have X not. And then we have half and v X squared minus V Nord squared and V f squared. So be right, the question finally as half Okay, except squad. Because extort squired zeros to that equation that turn will vanish. Bless mg mu K except ex notable Vanish on minus have m v North squared Because v I vf is going to vanish And that equals zero. As you can see that this equation is a Kodak trick equation off, you know, X squared plus bx plus c So we have this a x squared. Let's be X Bless this constant half envy nor squared plus c So we'll solve this good Adric equation. Bye. The formula X equals two minus B plus minus B squared minus for issue or two, eh? So therefore XF is equals to minus mg me. Okay. Bless minus square. Root off. MGM UK squared minus four times half okay, minus half And the north squad divided by do. Hey, so if you saw that, you will get minus mg mu k bless miners. Mg me. Okay, squad less. Okay. M v North squad divided by Kate So simplifying that we will get MGM UK over K. I'll just take it outside minus one bless minus square. Root off. Run! Bless Gay and V North squared over mg mu K squad. No, we can plug in the values in this case, you know, and you will see that you will get the answer off. 0.12 58 or zero point 13 meters. Hers. Now, for the be part you know, the block he has kept here and it is in the compressed state. So to remain are the compressed position in this case, the minimum coefficient of static friction us the magnitude of the force exerted by the spring. Okay, this should be equals to the coefficient of static friction times mg or the static friction force. So we don't want the block to move. So when the block is actually compressed to the maximum position and it comes to stop the force off, static friction is just sufficient. Two keep the block address. So for this condition, we have gay xf should be equals tomb us mg. If you saw that Mu s should be equals 2.77 now for the curd part of the problem, it says If the static friction is not large enough to hold the block in place, then the spring will push the block backwards towards equilibrium position. So we have the block kept here and the block is compressed in the B part. The static friction force were just enough to balance out the force of the spring. Or let's say, if it is not and the block will go towards equilibrium position and then the block with detach from bakery in position from their decompressing spring. So for this condition, you know, we again will write that we not is equals to zero and XF is equals to zero and we'll write the equation off the conservation of energy. Delta K Lester tie You bless force or friction times l should be equals to zero. So we have half mm v f squared minus V nodes squared less half gay x f squired minus x not squared. Bless mg me. Okay, X f miners. Ex North squad is equals to zero. So for the scenario will have half em V f squared because Venus squired is going to be zero or in this case, the block is initially addressed. So therefore vino disquiet as equals to zero. So we need to find out this final velocity for this condition minus half gay ex north squad. Because except for Z equals to zero victory in position Here, place mg mu k ex, not Dr Beak was zero. See if you saw that, then we should get VF. These equals two square root off. Gabe, I am ex not squad minus two g uk ex. Not so we can just ah very in this equation and plugging in the valleys of this equation off K as 1 20 100 per meter. And this is mass. A stupid Legrand's an ex, nor this 0.1258 G's 9.8 on mucous 0.3 and ex not is 11258 We get point 458 meters per second, which is approximately 0.5 meters for second. So that is our solution to the sea.

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