Solutions to Sample Problems 1. What is the force of friction between a block of ice that weighs 930 N and the ground if Mu = .12? (111.6 N) Here is what you start with:         W = 930 N Mu = .12 Once again, our formula for Mu is: F = Mu (FN) It's pretty easy to plug-in the stuff we know: F = (.12)(930 N) We then do the math to get 111.6 N. Go to: Problem Formulas Table of Contents 2. What is the coefficient of static friction if it takes 34 N of force to move a box that weighs 67 N? (.51) Here is what you start with:    F = 3.0 s      W = 67 N Once again, our formula for Mu is: F = Mu (FN) We know that FN is equal to W so we can plug what we know into the formula and get: 34 N = MuS (67 N) Divide the 34 by 67 and you end up with .507462 which happens to round to the answer of .51. Go to: Problem Formulas Table of Contents 3. A box takes 350 N to start moving when the coefficient of static friction is .35. What is the weight of the box? (1000 N) Here is what you start with:         F = 350 Mu = .35 Hey, since FN = W, surprisingly enough we use the formula: F = Mu (FN) which we plug-in the numbers and get 350 = .35 (FN) Divide 350 by .35 to get 1000 N. Go to: Problem Formulas Table of Contents 4. A car has a mass of 1020 Kg and has a coefficient of friction between the ground and its tires of .85. What force of friction can it exert on the ground? What is the maximum acceleration of this car? In what minimum distance could it stop from 27 m/s? (8500 N, 8.3 m/s/s, 43.8 m) Here is what you start with:         m = 1020 Kg Mu = .85 Use the formula W = mg with g = 9.8 m/s/s as the force of gravity. We now know the weight to be 9996 N. We can then use the formula F = Mu (FN) with FN equal to the weight. Plugged in it looks like: F = (.85)(9996 N) and equals 8496.6 N but with significant digits it ends up being 8500 N. In order to find the acceleration we need a formula that uses the Force we found and acceleration. Obviously F = ma seems best. We plug-in the numbers and get: 8496.6 N = (1020 Kg) a Solve for a and get 8.33 m/s/s as the acceleration But we're not done yet! We need to solve for the distance. Since this deals with good old velocity, acceleration and distance we need a formula that uses only what we know:            Vo = 37 m/s V = 0 (It comes to a complete stop) a = 8.33 m/s/s Now we can use: x = V2 - VO 2 ----------------2a x = 43.7575 or with sig figs, 43.8 m Go to: Problem Formulas Table of Contents 5. Clarice moves a 800 gram set of weights by applying a force of 1.2 N. What is the coefficient of friction? (.15) Here is what you start with:  m = 800 grams or .8 Kg           F = 1.2 N g = 9.8 m/s/s We first need to get the FN or Weight; we must convert the mass to kilograms and then get FN = 7.84 N for the next step. As usual when we look at the formulas we see that: F = Mu (FN) seems to be the only one that will uses Mu We plug-in the numbers to get: 1.2 = Mu (7.84 N) Divide 1.2 by 7.84 to get .15306 or .15 for the coefficient of friction. Go to: Problem Formulas Table of Contents 6. A car has a coefficient of friction between the ground and its tires of .85. What is the mass of the car if it takes 9620 N of force to make it slide along the ground? (1155 Kg) Here is what you start with:   Mu = .85       F = 9620 N We have to find the weight first so we use: F = Mu (FN) and that equals 9620 N = (.85)(FN) 9620 divided by .85 = 11317.647 N But we need this in mass so we use: F = mg 11317.647 = (m)(9.8 m/s/s) so the mass equals 1154.8619 or 1155 kg Go to: Problem Formulas Table of Contents 7. A 5.0 Kg block has a coefficient of friction of .15 on a flat surface. What is its acceleration if you exert a force of 15 N sideways on it when it is at rest? (Find the friction force first) (1.47 m/s/s) Here is what you start with:            m = 5.0 kg F = 15 N Mu = .15 So we can use the formula: FN = mg or 15 N = (5.0 kg)(9.8 m/s/s) So the FN = 49 N Then we plug that into: F = Mu (FN) that looks like F = (.15)(49 N) And get the actual force to be 7.35 N And then we can plug that back into the formula: F = ma or 7.35 N = (5.0)a Divide 7.35 by 5 and get 1.47 m/s/s Go to: Problem Formulas Table of Contents 8. A 10 Kg block is at rest on a level surface. It accelerates from rest to 51.2 m/s in 8 seconds when you exert a force of 100 N on it sideways. What is the acceleration of the block? What is the force of friction between the surface and the block, and what is the coefficient of friction? (6.4 m/s/s 36 N, .56) Here is what you start with:               m = 10 Kg Vo = 0 m/s V = 51.2 m/s t = 8 seconds F = 100 N    Find the acceleration using the our freindly neighborhood aceleration formula once again: V = Vo + at. Plug in a couple of numbers and get: 51.2 m/s = 0 m/s + (a * 8 s) Solve for a, and we get 6.4 m/s/s The force of friction = Force applied ñ The normal force FA = 100 N and FN = 64 N so the FF = 36 N Now we plug that back into the formula: FF = Mu (FN) we use our stuff 36 N = Mu (64 N) Divide 36 by 64 and get .5625 as the coefficient of friction. Go to: Problem Formulas Table of Contents 9. A 120 Kg log sled accelerates at 1.4 m/s/s when a horse pulls on it. What force must the horse exert if the coefficient of friction between the ground and the sled is .28? (497 N) Here is what you start with:  m = 120 Kg           Mu = .28 a = 1.4 m/s/s First we need to find FN and we use FN = mg F = (120 Kg)(9.8 m/s/s) and F equals 1176 N We then multiply that by the coefficient of friction (.28)to get the force of friction which is 329.28 N. We then use the formula: F = FA - FF, with the values plugged in we get: 168 = FA - 329.28 solve this little problem and get 497.28 or 497 N. Go to: Problem Formulas Table of Contents 10. (E.C. - Super Stud Problem) You exert a force of 24 N sideways on an object and it accelerates from 0 - 12 m/s over a distance of 5.2 m. You know that the coefficient of friction between the object and the ground is .58, so what is its mass? (1.23 Kg) Here is what you start with:          VO = 0 m/s V = 12 m/s Mu = .58         x = 5.2 m FA = 24 N Letís use x = V2 - VO2 ------------2a we plug in what we know and get: 5.2 m = 122 - 2 / 2a solve for acceleration and get a = 13.846 m/s/s we want to find F = ma but weíll need to use: F = FA - FF but we donít know the FF So FF = Mu * FN we plug stuff in and get: Since F = ma and FF = Mu * m * g we can put it into F = FA - FF so it looks like: ma = FA - (Mu * m * g) we can then put in what we know m (13.846) = 24 N ñ (.58 * m * 9.8m/s/s) Wait I donít Know! 13.846 = 24/m ñ 5.684 add 5.684m to both sides and NO That doesnít Work! 6.684m * (13.846) = 24 OHHH MAN! Ohhh dang it this just ends up being 1.23 Kg!!!!! Go to: Problem Formulas Table of Contents