What is the coefficient of static friction between the book and the tabletop?
Q. When you push a 1.85-kg book resting on a tabletop, it takes 2.22N to start the book sliding. Once it is sliding, however, it takes only 1.54 N to keep the book moving with constant speed. What is the coefficient of static friction between the book and the tabletop? What is the coefficient of kinetic friction between the book and the tabletop?
Asked by LostenPhysics - Sun Jun 14 14:38:47 2009 - - 2 Answers - 0 Comments
A. ues = Fs/mg = .122 uek = Fk/mg = .085
Answered by Steve - Sun Jun 14 14:44:19 2009
Q. When you push a 1.85-kg book resting on a tabletop, it takes 2.22N to start the book sliding. Once it is sliding, however, it takes only 1.54 N to keep the book moving with constant speed. What is the coefficient of static friction between the book and the tabletop? What is the coefficient of kinetic friction between the book and the tabletop?
Asked by LostenPhysics - Sun Jun 14 14:38:47 2009 - - 2 Answers - 0 Comments
A. ues = Fs/mg = .122 uek = Fk/mg = .085
Answered by Steve - Sun Jun 14 14:44:19 2009
Find a polynomial with integer coefficients and a leading coefficient of one that satisfies the given?
Q. Find a polynomial with integer coefficients and a leading coefficient of one that satisfies the given conditions. P has degree 2, and zeros 1 + i 5 and 1 - i 5. P(x) =?
Asked by Math - Tue Sep 22 16:54:24 2009 - - 1 Answers - 0 Comments
A. If x = a is a zero, then x - a is a factor. Since x = 1 + i 5 and x = 1 - i 5 are solutions, we have the following equation for P(x): P(x) = [x - (1 + i 5)][x - (1 - 5)] ==> P(x) = (x - 1 - i 5)(x - 1 + i 5) You can re-write this and get: P(x) = [(x - 1) + i 5][(x - 1) - i 5] This is a difference of squares, so we can get: P(x) = (x - 1) - (i 5) ==> P(x) = x - 2x + 1 - i (5) ==> P(x) = x - 2x + 1 - (-1)(5) ==> P(x) = x - 2x + 1 + 5 ==> P(x) = x - 2x + 6 Answer Verification: I hope that helps!
Answered by unknown - Tue Sep 22 20:24:30 2009
Q. Find a polynomial with integer coefficients and a leading coefficient of one that satisfies the given conditions. P has degree 2, and zeros 1 + i 5 and 1 - i 5. P(x) =?
Asked by Math - Tue Sep 22 16:54:24 2009 - - 1 Answers - 0 Comments
A. If x = a is a zero, then x - a is a factor. Since x = 1 + i 5 and x = 1 - i 5 are solutions, we have the following equation for P(x): P(x) = [x - (1 + i 5)][x - (1 - 5)] ==> P(x) = (x - 1 - i 5)(x - 1 + i 5) You can re-write this and get: P(x) = [(x - 1) + i 5][(x - 1) - i 5] This is a difference of squares, so we can get: P(x) = (x - 1) - (i 5) ==> P(x) = x - 2x + 1 - i (5) ==> P(x) = x - 2x + 1 - (-1)(5) ==> P(x) = x - 2x + 1 + 5 ==> P(x) = x - 2x + 6 Answer Verification: I hope that helps!
Answered by unknown - Tue Sep 22 20:24:30 2009
Find a polynomial with integer coefficients and a leading coefficient of one that satisfies?
Q. Find a polynomial with integer coefficients and a leading coefficient of one that satisfies the given conditions. Q has degree 3, and zeros -5 and 1 + i. Q(x) =?
Asked by Math - Tue Sep 22 16:23:09 2009 - - 1 Answers - 0 Comments
A. x^3+ kx^2 + m*x + c (x+5)(x-(1+i))(x-(1-i)) = (x+5)(x^2-x(1+i)-x(1-i)+( 1-i^2) = x^3 -2x^2 +2x +5x^2 -10x+ 10 = x^3+3x^2-8x+10 k=3, m=-8, c=10 Q(x)=x^3 + 3x^2 - 8x+ 10
Answered by unknown - Tue Sep 22 16:42:56 2009
Q. Find a polynomial with integer coefficients and a leading coefficient of one that satisfies the given conditions. Q has degree 3, and zeros -5 and 1 + i. Q(x) =?
Asked by Math - Tue Sep 22 16:23:09 2009 - - 1 Answers - 0 Comments
A. x^3+ kx^2 + m*x + c (x+5)(x-(1+i))(x-(1-i)) = (x+5)(x^2-x(1+i)-x(1-i)+( 1-i^2) = x^3 -2x^2 +2x +5x^2 -10x+ 10 = x^3+3x^2-8x+10 k=3, m=-8, c=10 Q(x)=x^3 + 3x^2 - 8x+ 10
Answered by unknown - Tue Sep 22 16:42:56 2009
What is the smallest coefficient of friction ever recorded?
Q. I am learning about coefficients friction, etc, in physics class (grade 11), and I was wondering if anyone knew what the smallest coefficient of friction is and between which two surfaces. I know that it has to be greater than 0, but I want to know what the smallest one is.
Asked by alamo49 - Sun Nov 8 20:22:09 2009 - - 1 Answers - 0 Comments
A. probably like oil against oil
Answered by philippohippo - Sun Nov 8 20:25:35 2009
Q. I am learning about coefficients friction, etc, in physics class (grade 11), and I was wondering if anyone knew what the smallest coefficient of friction is and between which two surfaces. I know that it has to be greater than 0, but I want to know what the smallest one is.
Asked by alamo49 - Sun Nov 8 20:22:09 2009 - - 1 Answers - 0 Comments
A. probably like oil against oil
Answered by philippohippo - Sun Nov 8 20:25:35 2009
What is the coefficient of static friction?
Q. A 310g paperback book rests on a 1.2kg textbook. A force is applied to the textbook, and the two books accelerate together from rest to 96cm in 0.42s . The textbook is then brought to a stop in 0.33s , during which time the paperback slides off. Within what range does the coefficient of static friction between the two books lie?
Asked by littlemonster - Thu Oct 1 10:27:12 2009 - - 1 Answers - 0 Comments
A. The frictional force is the one accelerating the paperback book. The frictional force required to accelerate the paperback book is then Fr1 = (0.31)* (0.92 / 0.42) = 0.67 N The frictional force required to de-accelerate the book to stop is then Fr2 = 0.31* (0.92 / 0.33) = 0.86 N I assumed that the velocity is 96 cm/s. The maximum static friction force is Fr max = us N = us mg From your problem, Fr1 < Fr max and Fr2 > Frmax, thus Fr1 < Frmax < Fr2 0.67 < Frmax < 0.86 or 0.67 < us mg < 0.86 dividing everything by mg we get 0.67/mg < us < 0.86/mg 0.67/(0.31*9.8) < us < 0.86/(0.31*9.8) 0.22 < us < 0.28 and that is your range for the static coefficient of friction us
Answered by Ivan A - Thu Oct 1 10:40:09 2009
Q. A 310g paperback book rests on a 1.2kg textbook. A force is applied to the textbook, and the two books accelerate together from rest to 96cm in 0.42s . The textbook is then brought to a stop in 0.33s , during which time the paperback slides off. Within what range does the coefficient of static friction between the two books lie?
Asked by littlemonster - Thu Oct 1 10:27:12 2009 - - 1 Answers - 0 Comments
A. The frictional force is the one accelerating the paperback book. The frictional force required to accelerate the paperback book is then Fr1 = (0.31)* (0.92 / 0.42) = 0.67 N The frictional force required to de-accelerate the book to stop is then Fr2 = 0.31* (0.92 / 0.33) = 0.86 N I assumed that the velocity is 96 cm/s. The maximum static friction force is Fr max = us N = us mg From your problem, Fr1 < Fr max and Fr2 > Frmax, thus Fr1 < Frmax < Fr2 0.67 < Frmax < 0.86 or 0.67 < us mg < 0.86 dividing everything by mg we get 0.67/mg < us < 0.86/mg 0.67/(0.31*9.8) < us < 0.86/(0.31*9.8) 0.22 < us < 0.28 and that is your range for the static coefficient of friction us
Answered by Ivan A - Thu Oct 1 10:40:09 2009
What is the coefficient of static friction between the crate and the incline?
Q. A 212-kg crate rests on a surface that is inclined above the horizontal at an angle of 20.4 . A horizontal force (magnitude = 531 N and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction between the crate and the incline?
Asked by Angela - Mon Oct 12 00:24:13 2009 - - 1 Answers - 0 Comments
A. Define: A = angle of the inclined plane = 20.4 m = mass of crate = 212 kg g = acceleration of gravity = 9.8m/s^2 Fa = Applied Force = 531 N W = weight of the crate = mg = 2078 N u = coefficient of static friction The forces on the crate parallel, Fp, and normal, Fn, to the surface is Fp = WsinA + FacosA Fn = WcosA - FasinA Fp is sufficient to begin the crate moving against the resistance from static friction. u = Fp/Fn = (WsinA + FacosA)/(WcosA - FasinA) = (2078sin20.4 + 531cos20.4)/(2078cos20.4 - 531sin20.4) u = .693
Answered by unknown - Mon Oct 12 00:42:52 2009
Q. A 212-kg crate rests on a surface that is inclined above the horizontal at an angle of 20.4 . A horizontal force (magnitude = 531 N and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction between the crate and the incline?
Asked by Angela - Mon Oct 12 00:24:13 2009 - - 1 Answers - 0 Comments
A. Define: A = angle of the inclined plane = 20.4 m = mass of crate = 212 kg g = acceleration of gravity = 9.8m/s^2 Fa = Applied Force = 531 N W = weight of the crate = mg = 2078 N u = coefficient of static friction The forces on the crate parallel, Fp, and normal, Fn, to the surface is Fp = WsinA + FacosA Fn = WcosA - FasinA Fp is sufficient to begin the crate moving against the resistance from static friction. u = Fp/Fn = (WsinA + FacosA)/(WcosA - FasinA) = (2078sin20.4 + 531cos20.4)/(2078cos20.4 - 531sin20.4) u = .693
Answered by unknown - Mon Oct 12 00:42:52 2009
What prevents a particle from slipping, the coefficient of kinetic or static friction?
Q. What prevents a particle from slipping, the coefficient of kinetic friction or the coefficient of static friction? And what are the inequalities to be aware of?
Asked by crazy girl - Sat Mar 28 23:04:16 2009 - - 1 Answers - 0 Comments
A. If it is still (not slipping), the coefficient of static friction applies. Force friction < normal force * coefficient
Answered by ( )Mistress Bekki - Sat Mar 28 23:09:16 2009
Q. What prevents a particle from slipping, the coefficient of kinetic friction or the coefficient of static friction? And what are the inequalities to be aware of?
Asked by crazy girl - Sat Mar 28 23:04:16 2009 - - 1 Answers - 0 Comments
A. If it is still (not slipping), the coefficient of static friction applies. Force friction < normal force * coefficient
Answered by ( )Mistress Bekki - Sat Mar 28 23:09:16 2009
How do you find coefficient of kinetic friction with only and angle and mass?
Q. A 15 kg book is placed on a 40o incline. The book is released and it slides down at constant speed. What is the coefficient of kinetic friction? Record your answer to two decimal places.
Asked by Kel P - Sat Nov 22 10:45:30 2008 - - 2 Answers - 0 Comments
A. Is it 0.38 to the second decimal place? prob wrong.
Answered by LMC - Sat Nov 22 10:54:24 2008
Q. A 15 kg book is placed on a 40o incline. The book is released and it slides down at constant speed. What is the coefficient of kinetic friction? Record your answer to two decimal places.
Asked by Kel P - Sat Nov 22 10:45:30 2008 - - 2 Answers - 0 Comments
A. Is it 0.38 to the second decimal place? prob wrong.
Answered by LMC - Sat Nov 22 10:54:24 2008
How to find the coefficient of kinetic friction?
Q. An object is accelerating down a plane that is inclined at an angle of 27.5 above the horizontal. The acceleration is 3.65 m/s. Find the coefficient of friction.
Asked by Josh Burns - Mon Nov 10 01:50:48 2008 - - 1 Answers - 0 Comments
A. I assume by 27.5 u mean 27.5 degrees. Coefficient of friction, M = F/N M = ma/[(mg)(sin b)] M = a/(gsin 27.5) M = 3.65/4.53 M = 0.806
Answered by Umagon - Tue Nov 11 14:14:59 2008
Q. An object is accelerating down a plane that is inclined at an angle of 27.5 above the horizontal. The acceleration is 3.65 m/s. Find the coefficient of friction.
Asked by Josh Burns - Mon Nov 10 01:50:48 2008 - - 1 Answers - 0 Comments
A. I assume by 27.5 u mean 27.5 degrees. Coefficient of friction, M = F/N M = ma/[(mg)(sin b)] M = a/(gsin 27.5) M = 3.65/4.53 M = 0.806
Answered by Umagon - Tue Nov 11 14:14:59 2008
Physics help! What is the coefficient of static friction between the sofa and the carpet?
Q. You need to move a 145-kg sofa to a different location in the room. It takes a force of 89 N to start it moving. What is the coefficient of static friction between the sofa and the carpet?
Asked by robert m - Mon Oct 26 17:44:31 2009 - - 1 Answers - 0 Comments
A. friction = u mg where u is the coefficient of friction we are told that you need 89N to get the sofa in motion; this is the force of friction needed to overcome, therefore we have 89=u (145x9.8) u=89/(145x9.8)=0.06
Answered by kuiperbelt2003 - Mon Oct 26 17:49:52 2009
Q. You need to move a 145-kg sofa to a different location in the room. It takes a force of 89 N to start it moving. What is the coefficient of static friction between the sofa and the carpet?
Asked by robert m - Mon Oct 26 17:44:31 2009 - - 1 Answers - 0 Comments
A. friction = u mg where u is the coefficient of friction we are told that you need 89N to get the sofa in motion; this is the force of friction needed to overcome, therefore we have 89=u (145x9.8) u=89/(145x9.8)=0.06
Answered by kuiperbelt2003 - Mon Oct 26 17:49:52 2009
ways to test the significance of a coefficient after doing a regression?
Q. I think we can use p-value to test the significance of the "Significance F" and each of the coefficient? on the other hand, we can also use T-Stat and F-stat to test the coefficient right? Or do we have to do both to increase the effectiveness of the argument? What if the P-Value of one coefficient is significant while the T-stat for that same coefficient is insignificant? How are we going to predict whether it is significant or not?
Asked by hetbh123 - Fri Oct 3 11:12:26 2008 - - 1 Answers - 0 Comments
A. t anf F are related by t^2=F. If you were using t, you would be using one formula and if you were using F, you would be using another formula. These formulas are designed so that t^2=F and when you use an F test, you'd use an F-table and when you use a t-test you would be using a t-table. The inference (whether the coefficients are significant or not) will be the same.
Answered by cidyah - Fri Oct 3 11:26:58 2008
Q. I think we can use p-value to test the significance of the "Significance F" and each of the coefficient? on the other hand, we can also use T-Stat and F-stat to test the coefficient right? Or do we have to do both to increase the effectiveness of the argument? What if the P-Value of one coefficient is significant while the T-stat for that same coefficient is insignificant? How are we going to predict whether it is significant or not?
Asked by hetbh123 - Fri Oct 3 11:12:26 2008 - - 1 Answers - 0 Comments
A. t anf F are related by t^2=F. If you were using t, you would be using one formula and if you were using F, you would be using another formula. These formulas are designed so that t^2=F and when you use an F test, you'd use an F-table and when you use a t-test you would be using a t-table. The inference (whether the coefficients are significant or not) will be the same.
Answered by cidyah - Fri Oct 3 11:26:58 2008
What coefficient is needed to balance the oxygen in the following equation?
Q. What coefficient is needed to balance the oxygen in the following equation? KClO3 ---> KCl + O2
Asked by Kia - Thu Aug 6 21:47:45 2009 - - 2 Answers - 0 Comments
A. 2KClO3 ---> 2KCl + 3O2 This is how you solve: Original equation is KClO3 ---> KCl + O2 K and Cl are already balanced. There are 3 O atoms on left and 2 O atoms on right. If you multiply O2 on the right by 3/2, then O atoms get balanced. So we write KClO3 ---> KCl + 3/2 O2 To remove fraction, multiply the complete equation by 2. We get 2KClO3 ---> 2KCl + 3O2 ___.
Answered by Somu - Thu Aug 6 21:57:15 2009
Q. What coefficient is needed to balance the oxygen in the following equation? KClO3 ---> KCl + O2
Asked by Kia - Thu Aug 6 21:47:45 2009 - - 2 Answers - 0 Comments
A. 2KClO3 ---> 2KCl + 3O2 This is how you solve: Original equation is KClO3 ---> KCl + O2 K and Cl are already balanced. There are 3 O atoms on left and 2 O atoms on right. If you multiply O2 on the right by 3/2, then O atoms get balanced. So we write KClO3 ---> KCl + 3/2 O2 To remove fraction, multiply the complete equation by 2. We get 2KClO3 ---> 2KCl + 3O2 ___.
Answered by Somu - Thu Aug 6 21:57:15 2009
How do I relate the coefficient of determination to extrapolation?
Q. I have data for a project I am doing, and I have set up the regression curves and the curve has a coefficient of determination (R^2) value of over .996. What I need to know is how to phrase my argument such that I prove the validity of any extrapolated prediction based on the extremly high value of the coefficient of determination. References not crucial, but if you think it'd help, go ahead.
Asked by theMayne - Fri Oct 19 00:51:54 2007 - - 1 Answers - 0 Comments
A. I seem to recall that the coefficient of determination is a value that explains what percentage of variation in the y-values of your analysis is due to it's linear relationship with the x-values. That is to say that when R^2 is approaching one, a great deal of the variation in your computed y-values is due to the linear relationship with the x-values. This can be used to say that using a linear model makes sense as the relationship is strongly linear due to the high R^2.
Answered by Modus Operandi - Sat Oct 20 00:35:16 2007
Q. I have data for a project I am doing, and I have set up the regression curves and the curve has a coefficient of determination (R^2) value of over .996. What I need to know is how to phrase my argument such that I prove the validity of any extrapolated prediction based on the extremly high value of the coefficient of determination. References not crucial, but if you think it'd help, go ahead.
Asked by theMayne - Fri Oct 19 00:51:54 2007 - - 1 Answers - 0 Comments
A. I seem to recall that the coefficient of determination is a value that explains what percentage of variation in the y-values of your analysis is due to it's linear relationship with the x-values. That is to say that when R^2 is approaching one, a great deal of the variation in your computed y-values is due to the linear relationship with the x-values. This can be used to say that using a linear model makes sense as the relationship is strongly linear due to the high R^2.
Answered by Modus Operandi - Sat Oct 20 00:35:16 2007
How to write a quartic equation with integral coefficient such that two of its roots are 8i and -2i?
Q. How to write a quartic equation with integral coefficient such that two of its roots are 8i and -2i?
Asked by Oops! - Wed Nov 5 21:10:41 2008 - - 1 Answers - 0 Comments
A. for any polynomial with real coefficients, if it has a complex root z, then it's complex conjugate z' is also a root. so if 8i and -2i are to be roots, -8i and 2i must also be. you then get a polynomial of the form: (x-8i)(x+8i)(x+2i)(x-2i), up to a multiplicative constant. now (x-8i)(x+8i) = x^2 + 64 and (x+2i)(x-2i) = x^2 + 4 so your polynomial is (x^2 + 64) (x^2 + 4) expand if needed. the coefficients will be integers, so you don't have to multiply by a constant; if the coefficients had been rational, you could have multiplied by a proper integer to get integral coefs.
Answered by Rich J - Thu Nov 6 00:02:22 2008
Q. How to write a quartic equation with integral coefficient such that two of its roots are 8i and -2i?
Asked by Oops! - Wed Nov 5 21:10:41 2008 - - 1 Answers - 0 Comments
A. for any polynomial with real coefficients, if it has a complex root z, then it's complex conjugate z' is also a root. so if 8i and -2i are to be roots, -8i and 2i must also be. you then get a polynomial of the form: (x-8i)(x+8i)(x+2i)(x-2i), up to a multiplicative constant. now (x-8i)(x+8i) = x^2 + 64 and (x+2i)(x-2i) = x^2 + 4 so your polynomial is (x^2 + 64) (x^2 + 4) expand if needed. the coefficients will be integers, so you don't have to multiply by a constant; if the coefficients had been rational, you could have multiplied by a proper integer to get integral coefs.
Answered by Rich J - Thu Nov 6 00:02:22 2008
Can the coefficient of static friction be less than the coefficient of kinetic friction?
Q. Can the coefficient of static friction be less than the coefficient of kinetic friction? Please help! I have a lab report due! Thank you so much!
Asked by darlingdaisy - Wed Jan 14 21:56:08 2009 - - 2 Answers - 0 Comments
A. In the overwhelming majority of cases, ues > uek Occasinally, ues = uek (teflon on teflon) The argument above mine is false because there is no law that says the motion of something should be easier to maintain than start.
Answered by Steve - Wed Jan 14 22:17:33 2009
Q. Can the coefficient of static friction be less than the coefficient of kinetic friction? Please help! I have a lab report due! Thank you so much!
Asked by darlingdaisy - Wed Jan 14 21:56:08 2009 - - 2 Answers - 0 Comments
A. In the overwhelming majority of cases, ues > uek Occasinally, ues = uek (teflon on teflon) The argument above mine is false because there is no law that says the motion of something should be easier to maintain than start.
Answered by Steve - Wed Jan 14 22:17:33 2009
What does it mean to have a negative coefficient in the regression model?
Q. What does it mean to have a negative coefficient in the regression model?
Asked by zedzo - Thu May 24 13:42:19 2007 - - 1 Answers - 0 Comments
A. It means that the variables are negatively associated. An increase in the independent variable leads to a decrease in the dependent variable. For instance if your equation was y = 4 - 3x, for every one unit increase in x, there is a 3 unit decrease in y.
Answered by Kelly - Thu May 24 13:47:20 2007
Q. What does it mean to have a negative coefficient in the regression model?
Asked by zedzo - Thu May 24 13:42:19 2007 - - 1 Answers - 0 Comments
A. It means that the variables are negatively associated. An increase in the independent variable leads to a decrease in the dependent variable. For instance if your equation was y = 4 - 3x, for every one unit increase in x, there is a 3 unit decrease in y.
Answered by Kelly - Thu May 24 13:47:20 2007
What minimum coefficient of static friction is needed if the car is to navigate the curve without slipping?
Q. A road with a radius of 73.8 m is banked so that a car can navigate the curve at a speed of 15 m/s without any friction. When a car is going 19.4 m/s on this curve, what minimum coefficient of static friction is needed if the car is to navigate the curve without slipping?
Asked by lhswolfette88 - Fri Sep 25 17:07:06 2009 - - 1 Answers - 0 Comments
A. (bank) = arctan(v^2/(rg)) = 17.2809205312689 deg, where v = 15 m/s The friction coefficient mu needed = total acceleration ATang along the banked surface divided by total acceleration ANorm normal to the banked surface. This is the difference of the centripetal (AcentTang) and gravitational (AgravTang) accelerations along the banked surface divided by the sum of the centripetal (AcentNorm) and gravitational (AgravNorm) accelerations normal to the banked surface. Now we set v = 19.4 m/s. Acent = v^2/r = 5.09972899728997 m/s^2 AcentTang = AcentCOS = 4.86952606895938 m/s^2 AcentNorm = AcentSIN = 1.51490980243883 m/s^2 Agrav = g = 9.8 m/s^2 AgravTang = AgravSIN = 2.91115784226766 m/s^2 AgravNorm = AgravCOS = 9.35762576818519 m/s^2 ANorm =… [cont.]
Answered by kirchwey - Fri Sep 25 18:06:22 2009
Q. A road with a radius of 73.8 m is banked so that a car can navigate the curve at a speed of 15 m/s without any friction. When a car is going 19.4 m/s on this curve, what minimum coefficient of static friction is needed if the car is to navigate the curve without slipping?
Asked by lhswolfette88 - Fri Sep 25 17:07:06 2009 - - 1 Answers - 0 Comments
A. (bank) = arctan(v^2/(rg)) = 17.2809205312689 deg, where v = 15 m/s The friction coefficient mu needed = total acceleration ATang along the banked surface divided by total acceleration ANorm normal to the banked surface. This is the difference of the centripetal (AcentTang) and gravitational (AgravTang) accelerations along the banked surface divided by the sum of the centripetal (AcentNorm) and gravitational (AgravNorm) accelerations normal to the banked surface. Now we set v = 19.4 m/s. Acent = v^2/r = 5.09972899728997 m/s^2 AcentTang = AcentCOS = 4.86952606895938 m/s^2 AcentNorm = AcentSIN = 1.51490980243883 m/s^2 Agrav = g = 9.8 m/s^2 AgravTang = AgravSIN = 2.91115784226766 m/s^2 AgravNorm = AgravCOS = 9.35762576818519 m/s^2 ANorm =… [cont.]
Answered by kirchwey - Fri Sep 25 18:06:22 2009
What is the coefficient of kinetic friction, uek between the block and the table?
Q. A block with mass m = 0.40 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m. When released, the block moves on a horizontal table top for 1.00 m before coming to rest. The spring constant k is 120 N/m. What is the coefficient of kinetic friction, uek between the block and the table?
Asked by Ashia C - Fri Mar 20 16:03:09 2009 - - 2 Answers - 0 Comments
A. Well remember that (Force of Kinetic Friction) = (friction constant)*(Normalforce) Normal in this case is just (m*g) as it must cancel out the weight the block applies to the table So Forcefriction = (mg)*friction constant Using consveration of energy we know that, U_spring - Work_friction = 0 (energy from beginning when at rest, to end when at rest again) Work_friction = Force of friction * distance_travelled = (mg)*force_constant*d_tra velled U_spring = (1/2)kx^2 (x being the distance compressed) with this you can easily solve for force_constant
Answered by mccragre - Fri Mar 20 16:15:32 2009
Q. A block with mass m = 0.40 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m. When released, the block moves on a horizontal table top for 1.00 m before coming to rest. The spring constant k is 120 N/m. What is the coefficient of kinetic friction, uek between the block and the table?
Asked by Ashia C - Fri Mar 20 16:03:09 2009 - - 2 Answers - 0 Comments
A. Well remember that (Force of Kinetic Friction) = (friction constant)*(Normalforce) Normal in this case is just (m*g) as it must cancel out the weight the block applies to the table So Forcefriction = (mg)*friction constant Using consveration of energy we know that, U_spring - Work_friction = 0 (energy from beginning when at rest, to end when at rest again) Work_friction = Force of friction * distance_travelled = (mg)*force_constant*d_tra velled U_spring = (1/2)kx^2 (x being the distance compressed) with this you can easily solve for force_constant
Answered by mccragre - Fri Mar 20 16:15:32 2009
What is the coefficient of kinetic friction between the object and the floor?
Q. An object slides on a level surface in the +x direction. It slows and comes to a stop with a constant acceleration of -2.45 m/s^2. What is the coefficient of kinetic friction between the object and the floor?
Asked by loving life - Tue Aug 19 11:36:21 2008 - - 2 Answers - 0 Comments
A. ue = a/g = 2.45/9.8 = .25
Answered by Steve - Tue Aug 19 12:04:09 2008
Q. An object slides on a level surface in the +x direction. It slows and comes to a stop with a constant acceleration of -2.45 m/s^2. What is the coefficient of kinetic friction between the object and the floor?
Asked by loving life - Tue Aug 19 11:36:21 2008 - - 2 Answers - 0 Comments
A. ue = a/g = 2.45/9.8 = .25
Answered by Steve - Tue Aug 19 12:04:09 2008
what is the coefficient of sliding friction between furniture and the floor?
Q. An applied force of 365N is applied to drag a furniture piece across the floor with an acceleration of 1.50m/s^2. If the mass is 65.0kg, what is the coefficient of sliding friction between furniture and the floor? Thanks for all your help.
Asked by Anne B - Wed Oct 29 11:05:52 2008 - - 2 Answers - 0 Comments
A. we have F = ma F = 65 * 1.5 = 97.5 N this is the force required if there were no friction. Applied force = 365 N thus Opposing force [or the frictional force] = 365 - 97.5 = 267.5 Now, if the coefficient of friction is u, then, u = f/R where, f = frictional force and R = normal reaction = m * g = 65 * 9.8 = 637 thus, u = 267.5 / 637 = 0.4199 or 0.42 thus, coefficient of friction is, 0.42
Answered by Mak - Wed Oct 29 11:13:56 2008
Q. An applied force of 365N is applied to drag a furniture piece across the floor with an acceleration of 1.50m/s^2. If the mass is 65.0kg, what is the coefficient of sliding friction between furniture and the floor? Thanks for all your help.
Asked by Anne B - Wed Oct 29 11:05:52 2008 - - 2 Answers - 0 Comments
A. we have F = ma F = 65 * 1.5 = 97.5 N this is the force required if there were no friction. Applied force = 365 N thus Opposing force [or the frictional force] = 365 - 97.5 = 267.5 Now, if the coefficient of friction is u, then, u = f/R where, f = frictional force and R = normal reaction = m * g = 65 * 9.8 = 637 thus, u = 267.5 / 637 = 0.4199 or 0.42 thus, coefficient of friction is, 0.42
Answered by Mak - Wed Oct 29 11:13:56 2008
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