5

A block of mass 2kg is pulled at constant velocity a long a rough horizontal surface by an applied force Fz4ON, Then the frictional force is: (2 Points)30"m17,...

Question

A block of mass 2kg is pulled at constant velocity a long a rough horizontal surface by an applied force Fz4ON, Then the frictional force is: (2 Points)30"m17,3225.9843.3034.64

A block of mass 2kg is pulled at constant velocity a long a rough horizontal surface by an applied force Fz4ON, Then the frictional force is: (2 Points) 30" m 17,32 25.98 43.30 34.64



Answers

A $2 \mathrm{~kg}$ block is placed over a $4 \mathrm{~kg}$ block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0 -20. Find the acceleration of the two blocks if a horizontal force of $12 \mathrm{~N}$ is applied to (a) the upper block, (b) the lower block. Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$

So we'll go back to the ocean. We have we can applaud visits moving visit uh pulled by the force that is equal to 61 it falls P equals two in the water. It's like. So basically uh we need to find that and the mass of the block that is also given that is equal to 2.5 K. So we need to find that. We need to find. We need to find little mind though the pigs and forced into the direction of friction force that's magnitude and direction of the picture from when P equals stupid. New 10 10 10 and 12 inches right, nitrogen 10 billion to life. And and the coefficient of static friction. A geo 100.2 and protections and 0.25 So new as equal to 0.4 and milk equal to zero point 25 Right? So in the first part, what you can do over here is that massive? 2.5 kg. What will the MG MGB even think 25 MG if you're taking the 10. So 25. So if P equals two in this personally of which we have given that if P. Is equal to urgently going everyone's this forces, then what would be the normal force? The normal force in this case will be nothing. Uh 17. You can. Right thanks. So now what we need to fix in post? Mm hmm. So in this case The friction force will be obviously in the town, it is direction. And what will the value? So in this year's fix and force that will be moved. Times of numerous times of this normal 17 times of European four The 6.8. And it is exceeding the value of the pulling force. That means uh body Egypt dress. So that means the fix and possibly only act. That is equal to the fourth pool. That wins the fr Will be just six million because they're exceeding the value of the pulling force. And in the next part, When this p equal to 10, you can I ask. Mhm. Uh it is equal to 10. And then what when your natural force then known and false equals to uh pretending to now Not fr will be nothing but a 0.4 into 15. That means six Children. And there it was just about to see it. But it is equal to 16 6 newton. Right? And the next part, When P equals two equals to 212 illusion, that means the normal force and only 25 minutes to 13. So now F. R. Equals 2. 13 into 0.4. That must happen to. That means it is moving now now it is moving. So what we need to fix and personal, you need to use the kinetic kinetic friction coefficient so far will be 13 into zero point of 25 So if you can direct one by four as well. So this will be three point 3.3 point 3.1002 and five. New channel. Thank you. And obviously the Budget Director.

So here we can apply Newton's second law to the X in my direction. So the horizontal force minus the magnitude of the frictional force equaling the mass times acceleration in the extraction. Here, the vertical force plus the normal force minus the gravitational force equals zero. We're here F this equaling 6.0 Newton's and the Mass is gonna be equaling 22.5 kilograms. So we can say that four part A, Given that the vertical forces equaling 8.0 Newton's, we can say that the normal force is I'm gonna be equaling to 2.5 kilograms multiplied by 9.8 meters per second squared and there's gonna be minus 8.0 Newton's and this is gonna give us 16 0.5 Newton's. Now we can say that the maximum the maximum static frictional force is going to be equaling to the coefficient of static friction multiplied by the normal force. This would be equaling 2.40 multiplied by 16.5 Newton's. This is equaling 26.6 Newtons, which is greater than F of 6.0 Newton's. So here block doesn't move acceleration of zero and we can say that. Then the force, the frictional forces equaling the magnitude of the the magnitude of the the vertical force. My apologies. And this is equaling 8.0 mutants. So we can then say that he rather my college is six point. And so here we can then say that for part B P is equaling 10 Newtons and we can say that then the normal force is gonna be equaling two 14.5 Newtons. The Frictional Force, the maximum static frictional force will be equaling two 0.40 multiplied by 14.5 Newtons. And this is giving us 5.8 Newtons, which is actually less than F so here, given that it's at less than half of 6.0 Newton's, we can say that here the block does move And here the Canada, the friction will be going to the kinetic friction equaling 0.25 multiplied by the normal force of 14.5 Newton's. And this is giving us 3.6 Newton's for part C, then P equaling 12. Newton's then leads to the normal force equaling 12.5 Newtons, which is giving a maximum static frictional force equaling 0.40 multiplied by 12.5 Newton's. This is giving us five brother, 5.0 Newton, which is, of course, less than half of 6.0 Newton's. And so the forced the frictional forces equaling the kinetic frictional force of point to five multiplied by 12.5 new tins. And this is giving us approximately 3.1 Newton's. This would be our answer for part C. That is the end of the solution. Thank you for watching.

In this question, we're trying to determine the force of friction acting on a block. So first let's draw our free body diagram for our situation. So we gotta force after the right the force of gravity down force of friction opposite that force. And then up we have some force P. And our normal force. Yeah. Okay. So what we can say in this situation up and down has to be balanced to the block is and flying up into the air, breaking down to the table. So please p. Plus the normal force is equal to the force of gravity and left and right. We don't know if they're balanced or unbalanced yet but we really don't care. All we know is that this force is six mm. So we want to figure out the force of friction. So a couple other pieces of information that we need. So the mass of the block Is 2.5 kg. The coefficient of static friction is .4. The coalition of kinetic friction This .25. Mhm. Okay. Yeah. So now we are using three different values for P. To determine our normal force. Let's move this equation around here, I'm gonna subtract peter both sides. So the normal force is equal to F. G minus P. Yeah. Yeah. Mhm. Okay, so let's calculate the force of gravity first. So force of gravity equals 2.5 kg. The force of gravity is mass times gravity Times 9.8 meters per second squared. So our force of gravity acting on the block is always going to be 24.5 mittens. Okay, so now let's calculate our three normal forces for the three situations. So in part A P equals eight newtons. So when we plug into our equation here, F.G. This is going to give me a normal force. Uh 16.5 Newtons in B. He is tendons. This gives me a normal force Of 14 5 Newtons. And see here's 12 movements, which gives me a normal force of 12.5 mittens. Okay, so now how do we figure out the forces of friction? So in each part we'll start with part A. We could have the force of friction static which is less than or equal to the coefficient of static friction comes normal force or the force of friction, kinetic is equal to the coalition of kinetic friction times normal force. So the way that we're going to know which one to use is we're going to start with static friction. Okay. And if static friction, when we calculate here, if the force of friction is less than or equal to a number that's bigger than six because that's our force here, We know it's not going to move. Static friction has to be less than six in order for the back to block to be moving. So let's first calculate for part A. So we plug in we get four sub static friction is less than equal to 0.4 and our normal force which is 16.5 Millions. So this comes out to force of friction is equal to Yeah. All right. Not equal to his less than equal to 6.6. Newton's Okay. So since static friction can be bigger then are pulling force of F. That means that the force of static friction is only going to balance it out. So our static friction in part A. Is six newtons. We don't have to worry about carrie correction. So part of our friction is going to be six. Newtons just exactly balancing out are pulling force of F. So part B. Same thing. We're gonna start with static friction less than equal to 0.4 times 14.5 Newtons. And this comes out to remember Yeah, Which only comes out to 5.8. So, since our number is less than six, we have to switch over to kinetic friction. So when we plug in for kinetic friction, That's .25 times are 145. Newton number force. So our force of friction kinetic In this scenario is going to be 3625, which will earn a 36 newtons. Mhm. And part C. Mm. Well, since our force of friction was less than six in part B, it's going to be even less than that. So, we can skip the force of static friction in this part and go right to kinetic. So .25, 10- 12.5 Newtons. And this comes out to are force of friction. It is 3.1 newtons. So in part A it's static friction exactly equal to six. The block is not moving at all, and then part B and C. It's kinetic friction, and we have our two values here.

For this problem. On the topic of force and motion, we are told that a 2.5 kg block is addressed on a horizontal surface. A force F of magnitude six Newtons acts on the block horizontally. And a vertical force pr then applied to the block has shown in the figure. The coefficient of friction for the block and surface are U. S. 0.4 and U k 0.25 We want to find the magnitude of the frictional force acting on the block. If he has a magnitude of eight newtons, B, 10 newtons and c 12 newtons. Now we can choose X to be positive right words, and why it be positive upwards. And we can apply Newton's second law in the horizontal direction. We have the applied force F minus the frictional force. Little F. Is equal to M. A. In the vertical direction. We have P plus the normal force that the flow exerts on the block, minus the weight of the block. Mg was equal to zero. Since there's no net motion in the vertical election, we have F. Being six newtons and M 2.5 kg. And so for part A. If P is equal 28 newton's, then we can see that the noble force F. N. Is equal to 2.5 kg times 9.8 m per square second minus P, which is eight newton's. And so the normal force has magnitude 16.5 newtons. And so the static frictional force Fs max is equal to us times F. N, which is 6.6 newtons. And so this is larger than the six newton right would force so the block does not move. Which so let the acceleration be zero in the and into the first. For equations above, we get the frictional force F to equal to the blood force P, which is six newton's in part B. We have the applied force P. To be 10 newtons, which gives us the normal force F. N. To be 2.5 kg times 9.8 m per square second. And so the normal force and this is minus P, which is minus 10 m. So the noble force is 14.5 newtons. And the maximum static frictional force F. S. Max in this case, new S. Times F. N. Is 5.8 newtons. This is less than the six nutrient right would force. So the block block does move. So we're not dealing with static but we're dealing with kinetic friction. And so the kinetic frictional force F. K. Is equal to the coefficient of clinic fiction mu K times the normal force FM. Which gives us a frictional force of 3.6 newtons. For part C. We have the applied force p equal to 12 newtons. This gives us the normal force F. N. To be 12 0.5 newtons. And so the maximum static frictional force is equal to five newtons, which is less than the six newton Reitman force. And so again the block moves. And we have kinetic friction, and so the kinetic frictional force f. K, is equal to the coefficient of kinetic friction times This normal force FN, which gives us the frictional force, the 0.1 newtons.


Similar Solved Questions

5 answers
9: Problem 7PreviousProblem ListNextpoint)Find the work done by the force field F(x,Y, 2) = 6xI + 6yJ + 4K on a particle that moves along the helix X(t) = 5 cos(t)I + 5 sin(t)J + 6tK,0 < t < 2T .
9: Problem 7 Previous Problem List Next point) Find the work done by the force field F(x,Y, 2) = 6xI + 6yJ + 4K on a particle that moves along the helix X(t) = 5 cos(t)I + 5 sin(t)J + 6tK,0 < t < 2T ....
5 answers
Suppose that Xn is a sequence of real numbers that converges to 1 as n 4 C Prove that the following limit exists: 1 + 2xu 47 3 as n 57 C
Suppose that Xn is a sequence of real numbers that converges to 1 as n 4 C Prove that the following limit exists: 1 + 2xu 47 3 as n 57 C...
5 answers
(20 Points) Evaluate SzF dr where F (y,x2) and C is given by r(t) (4 -+,4t _ 02) for 0 <t<3.
(20 Points) Evaluate SzF dr where F (y,x2) and C is given by r(t) (4 -+,4t _ 02) for 0 <t<3....
5 answers
A simple RC circuit is shown below_ where R SOMQ2 and € = LSOuF. The capacitor has an initial charge of 90.OpC before the switch is elosed determine the time constant of the circuit b) determine the charge on the capacitor 3.0Os afier the switch closes determine the current in the circuit 4.S0s after the switch closesswitch
A simple RC circuit is shown below_ where R SOMQ2 and € = LSOuF. The capacitor has an initial charge of 90.OpC before the switch is elosed determine the time constant of the circuit b) determine the charge on the capacitor 3.0Os afier the switch closes determine the current in the circuit 4.S0...
5 answers
Problem(19) 8 points) Let be the solid rcgion and the thrcc coordinatc plancs and the plane 3c + 3y + 2 = 9 and let F(z, y, 2) = 2ri + (y2)j + 3zk . FindF. NdS
Problem(19) 8 points) Let be the solid rcgion and the thrcc coordinatc plancs and the plane 3c + 3y + 2 = 9 and let F(z, y, 2) = 2ri + (y2)j + 3zk . Find F. NdS...
5 answers
An 80-kg patient went to a doctor. The doctor advised him to reduce his weight. Should the patient plan to live on the moon so that his weight gets reduced? Explain your answer.
An 80-kg patient went to a doctor. The doctor advised him to reduce his weight. Should the patient plan to live on the moon so that his weight gets reduced? Explain your answer....
4 answers
Find basis fr the column space or 4
Find basis fr the column space or 4...
1 answers
Establish each identity. $$ \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}+1=2 \cos ^{2} \theta $$
Establish each identity. $$ \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}+1=2 \cos ^{2} \theta $$...
5 answers
Internet sites that allow people to post their resumes reduce the cost of a job search. How do you think the Internet has affected the natural rate of unemployment?
Internet sites that allow people to post their resumes reduce the cost of a job search. How do you think the Internet has affected the natural rate of unemployment?...
5 answers
1Calculate the enthalpy change for the following reaction FEOls) _ FezOsls) FeaOals) Given the following information (in kJ) 2Fe{s} Ozlg) 2FcOls) ZHe _ =-545 2FeOls) #Ozlg) + FezOa(s) oH? 828- Fefs) 2Ozlg) FesOals) 'oHz =-11110.5
1 Calculate the enthalpy change for the following reaction FEOls) _ FezOsls) FeaOals) Given the following information (in kJ) 2Fe{s} Ozlg) 2FcOls) ZHe _ =-545 2FeOls) #Ozlg) + FezOa(s) oH? 828- Fefs) 2Ozlg) FesOals) 'oHz =-111 10.5...
5 answers
Begin by graphing $f(x)=2^{x} .$ Then use transformations of this graph and a table of coordinates to graph the given function. If applicable, use a graphing utility to confirm your hand-drawn graphs.$g(x)=2^{x+1}$
Begin by graphing $f(x)=2^{x} .$ Then use transformations of this graph and a table of coordinates to graph the given function. If applicable, use a graphing utility to confirm your hand-drawn graphs. $g(x)=2^{x+1}$...
5 answers
The specific heat capacity of methane gas is 2.094 J/g-K. Howmany kilojoules of heat are needed to raise the temperature of 58.7g of methane from 30.3 °C to 60.7 °C?
The specific heat capacity of methane gas is 2.094 J/g-K. How many kilojoules of heat are needed to raise the temperature of 58.7 g of methane from 30.3 °C to 60.7 °C?...
5 answers
An aerobic reaction is one that requiresQuestion 302ptsis the rate of energy use for metabolism under basal conditions, usually expressed as kcalories per kilogram body weight per hour:Question 312 ptsAn individual with a BMI of 'greater or 100 poundsor more overwelght Is categorized as clinically severe obese.
An aerobic reaction is one that requires Question 30 2pts is the rate of energy use for metabolism under basal conditions, usually expressed as kcalories per kilogram body weight per hour: Question 31 2 pts An individual with a BMI of 'greater or 100 poundsor more overwelght Is categorized as c...
4 answers
Fnd two sets of polar coordinates for the point for 0 _ 0 < 0)8.(smaller rvalue)(, 0)8.(larger rvalue)Need Help?EE#EUHLDHSHALULZ]
Fnd two sets of polar coordinates for the point for 0 _ 0 < 0) 8. (smaller rvalue) (, 0) 8. (larger rvalue) Need Help? EE#EUHLD HSHALULZ]...

-- 0.019627--