Change the chapter
A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake as shown in Figure 5.21(a). (a) Calculate the minimum force F he must exert to get the block moving. (b) What is the magnitude of its acceleration once it starts to move, if that force is maintained?
Question Image
<b>Figure 5.21</b> A person (a) pushing, and (b) pulling, a block of ice across ice.
Figure 5.21 A person (a) pushing, and (b) pulling, a block of ice across ice.
Question by OpenStax is licensed under CC BY 4.0.
Final Answer
  1. $50\textrm{ N}$
  2. $0.7 \textrm{ m/s}^2$
Solution Video

OpenStax College Physics Solution, Chapter 5, Problem 18 (Problems & Exercises) (6:56)

Sign up to view this solution video!


3 votes with an average rating of 3.7.

Calculator Screenshots

OpenStax College Physics, Chapter 5, Problem 18 (PE) calculator screenshot 1
Video Transcript

This is College Physics Answers with Shaun Dychko. This person is pushing a block of ice on ice and we can find the coefficient's of static and kinetic friction from table [5.1]; static friction coefficient is 0.1 and the kinetic friction coefficient once it starts sliding is 0.03 and we are asked what minimum force do they need to apply in order to get the block just to begin moving? And so the type of friction involved there is static friction because the block is not yet moving. Now notice that the person is pushing down at this angle of 25 degrees with respect to horizontal. Doing so means that they are actually increasing the force downwards on the block... they are adding to the gravity because there's some y-component to their applied force and that in turn causes an increase in the normal force upwards and by increasing the normal force, they are increasing the static friction force. So we are going to consider the y-direction here and we'll figure out an expression that will involve this normal force that we will then use in our consideration of the horizontal direction, the x-direction, which we'll do in the second step. So we have written down all the things that we know of course: the mass being 45 kilograms, okay. In the y-direction, Newton's second law says that the normal force upwards minus all the other forces downwards has to equal mass times acceleration but there's no acceleration vertically and so we'll just say 0 here. And then we can make substitution's for each of these terms here: we have gravity is mass of the block times <i>g</i> and then the y-component of this person's force is the force multiplied by <i>sin</i> of 25 degrees because this y-component is the opposite leg of this this right triangle here. So we take <i>sin</i> of the angle, multiply it by the hypotenuse to get the opposite leg and then we make substitution's for both those terms and we call this equation number 1. So this is the normal force upwards minus gravity downwards minus the component of the applied force downwards equals zero. And then considering the x-direction, we have that the x-component of the applied force by the person to the right so it's positive minus the friction force to the left all equals zero. This is just the border between when there is acceleration or is not and we are using the maximum static friction force here. So the x-component of the applied force is the force multiplied by <i>cos Θ</i> because we are finding the adjacent leg of this right triangle now and multiplying <i>cos Θ</i> by the hypotenuse to get the x-component. And then the maximum static friction force is equal to the coefficient of static friction multiplied by the normal force and so we make substitution's for each of these terms using what we have written here and we'll call this equation number 2. So we have <i>Fcos Θ</i> minus <i>μ sF N</i> equals 0. So we have two equations and there are a couple of things that we don't know but we can make a substitution from equation 1 and solve it instead for <i>F N</i> and plug that in up here and then we'll be able to figure out what this applied force needs to be. So here's equation 1 written again but version b because we rearranged it a bit: we have added <i>mg</i> to both sides and added <i>Fsin Θ</i> to both sides and then we end up with normal force is <i>mg</i> plus <i>Fsin Θ</i>. And so then we rewrite equation 2, we'll call it version b where we have made a substitution for the normal force and written <i>mg</i> plus <i>Fsin Θ</i> in its place. So we have <i>Fcos Θ</i> minus <i>μ s</i> times <i>mg</i> plus <i>Fsin Θ</i> equals 0 then distribute the coefficient of static friction into the bracket and then we end up with this line and we are solving for <i>F</i> remember so let's collect the two terms that contain a factor <i>F</i> together and then factor out the <i>F</i> so we have <i>F</i> times <i>cos Θ</i> minus <i>μ ssin Θ</i> and then take the other term to the right side so we'll add <i>μ smg</i> to both sides. Okay and then we can divide both sides by this bracket and we'll solve for <i>F</i>. So we have <i>F</i> is <i>μ smg</i> divided by <i>cos Θ</i> minus <i>μ ssin Θ</i>. So that's 0.1—which is the coefficient of static friction of ice on ice— times 45.0 kilograms—mass of the block— times 9.80 meters per second squared divided by <i>cos</i> of 25 degrees minus 0.1 times <i>sin</i> 25 which is 50 newtons. Then in part (b), we are asked well, given that this applied force remains the same after it starts moving, what will its acceleration be when it's moving? So we are going to look at equation 2 again; here's equation 2: <i>Fcos Θ</i> minus <i>μF N</i> equals 0 but the <i>μ</i> is gonna be substituted with the coefficient of kinetic friction now because the block is sliding and it's going to be accelerating so we don't have a 0 here anymore, we have an <i>ma</i> here now—this is Newton's second law— and our job is to solve for <i>a</i> so we divide both sides by <i>m</i> and we have the acceleration is <i>Fcos Θ</i> minus <i>μ KF N</i> over <i>m</i> but this normal force, we need to get rid of that by substituting it with this expression— <i>mg</i> plus <i>Fsin Θ</i>— and so we make that substitution here in place of <i>F N</i> and then at this point, you can plug in numbers I decided to make it a bit more clean looking by factoring out the <i>F</i> between this term and this one after multiplying by the negative <i>μ K</i> so that's why there's a minus there because it's being multiplied by this minus <i>μ K</i>. So we have <i>F</i> times <i>cos Θ</i> minus <i>μ Ksin Θ</i> minus <i>μ Kmg</i> all over <i>m</i>. So that's 51.039 newtons— using the unrounded answer from part (a); part (a) having only one significant figure because this coefficient of static friction has only one significant figure that's why I wrote 50 newtons but the unrounded answer is 51.039 newtons times <i>cos</i> 25 degrees minus 0.03— coefficient of kinetic friction— times <i>sin</i> 25 minus 0.03 times 45.0 kilograms times 9.80 meters per seconds squared divided by the mass gives us 0.7 meters per second squared.


Submitted by raynellmcclellan on Tue, 05/19/2020 - 11:47

Why do given the solution as 50 if it comes out as 51. Are we suppose to be rounding to the nearest tenth?

Submitted by ShaunDychko on Wed, 05/20/2020 - 15:43

Yes, exactly, we're rounding to one significant figure. This is due to the of the coefficient of static friction, which has only one significant figure. When multiplying by a number with one sig. fig., the answer gets only one sig. fig. also.
All the best,

In reply to by raynellmcclellan