g and then the y-component that this person's force is the pressure multiplied by sin that 25 degrees due to the fact that this y-component is the opposite foot of this this best triangle here. So us take sin the the angle, main point it by the hypotenuse to get the the opposite leg and then we make substitution's because that both those terms and also we speak to this equation number 1. For this reason this is the normal force upwards minus gravity downwards minus the ingredient of the applied force downwards equals zero. And then considering the x-direction, we have actually that the x-component the the applied force through the person to the right so it's optimistic minus the friction pressure to the left all equals zero. This is simply the border between when over there is acceleration or is not and we space using the maximum static friction pressure here. Therefore the x-component of the applied force is the force multiplied by cos Θ due to the fact that we room finding the surrounding leg that this best triangle now and also multiplying cos Θ by the hypotenuse to obtain the x-component. And then the maximum revolution friction pressure is equal to the coefficient of revolution friction multiplied by the normal force and also so we make substitution's because that each of these terms making use of what we have written here and we'll call this equation number 2. So we have actually Fcos Θ minus μ sF N equals 0. Therefore we have actually two equations and there are a pair of things that us don't know yet we deserve to make a substitution from equation 1 and solve it instead for F N and also plug that in increase here and also then we'll be able to figure the end what this applied force demands to be. For this reason here's equation 1 written again but version b since we rearranged the a bit: we have added mg to both sides and added Fsin Θ to both sides and also then we finish up v normal pressure is mg to add Fsin Θ. And so then us rewrite equation 2, we'll call it variation b wherein we have actually made a substitution because that the typical force and also written mg add to Fsin Θ in the place. For this reason we have actually Fcos Θ minus μ s time mg to add Fsin Θ equals 0 climate distribute the coefficient of revolution friction into the bracket and also then we finish up through this line and we are addressing for F remember therefore let's collection the two terms that contain a aspect F together and also then factor out the F for this reason we have F times cos Θ minus μ ssin Θ and then take the other term come the best side therefore we'll include μ smg to both sides. Okay and also then we deserve to divide both sides by this bracket and also we'll solve for F. So we have F is μ smg separated by cos Θ minus μ ssin Θ. Therefore that's 0.1—which is the coefficient of revolution friction of ice cream on ice— time 45.0 kilograms—mass that the block— times 9.80 meter per 2nd squared separated by cos the 25 levels minus 0.1 times sin 25 i m sorry is 50 newtons. Then in part (b), we are asked well, provided that this applied force continues to be the same after that starts moving, what will certainly its acceleration be when it's moving? for this reason we space going to look in ~ equation 2 again; here's equation 2: Fcos Θ minus μF N amounts to 0 but the μ is gonna be substituted with the coefficient of kinetic friction now because the block is sliding and also it's going come be speeding up so we don't have actually a 0 here anymore, we have an ma below now—this is Newton's 2nd law— and also our project is to settle for a for this reason we divide both sides by m and also we have the acceleration is Fcos Θ minus μ KF N over m yet this common force, we require to eliminate that by substituting it v this expression— mg to add Fsin Θ— and also so we make the substitution right here in ar of F N and then in ~ this point, you can plug in numbers I determined to do it a bit much more clean spring by factoring out the F between this term and this one after multiplying by the an adverse μ K therefore that's why there's a minus there because it's gift multiplied by this minus μ K. So we have F times cos Θ minus μ Ksin Θ minus μ Kmg almost everywhere m. For this reason that's 51.039 newtons— utilizing the unrounded answer from part (a); component (a) having only one significant figure since this coefficient of revolution friction has actually only one far-reaching figure that's why I created 50 newtons but the unrounded answer is 51.039 newtons time cos 25 degrees minus 0.03— coefficient the kinetic friction— times sin 25 minus 0.03 times 45.0 kilograms time 9.80 meter per seconds squared divided by the mass offers us 0.7 meters per 2nd squared.">

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This is university Physics Answers v Shaun Dychko. This human being is pushing a block of ice on ice and also we can uncover the coefficient"s the static and also kinetic friction from table <5.1>; static friction coefficient is 0.1 and also the kinetic friction coefficient as soon as it starts sliding is 0.03 and also we room asked what minimum force do they need to apply in bespeak to acquire the block simply to start moving? and so the form of friction connected there is static friction due to the fact that the block is not yet moving. Now notification that the human being is pushing under at this angle of 25 levels with respect to horizontal. Act so means that they are actually enhancing the pressure downwards ~ above the block... Lock are including to the gravity since there"s some y-component come their used force and that subsequently causes rise in the normal force upwards and by boosting the normal force, lock are increasing the static friction force. Therefore we space going to consider the y-direction here and also we"ll figure out an expression that will certainly involve this normal pressure that we will then use in our factor to consider of the horizontal direction, the x-direction, which we"ll execute in the second step. So we have written under all the points that we recognize of course: the mass gift 45 kilograms, okay. In the y-direction, Newton"s second law states that the normal pressure upwards minus all the other forces downwards has to equal mass time acceleration but there"s no acceleration vertically and also so we"ll just say 0 here. And also then we have the right to make substitution"s because that each of this terms here: we have actually gravity is massive of the block time g and also then the y-component the this person"s force is the pressure multiplied by sin of 25 degrees because this y-component is the opposite leg of this this best triangle here. So we take sin the the angle, main point it through the hypotenuse to obtain the opposite leg and also then we make substitution"s for both those terms and we speak to this equation number 1. So this is the normal force upwards minus gravity downwards minus the component of the used force downwards amounts to zero. And then considering the x-direction, we have that the x-component that the used force by the person to the right so it"s hopeful minus the friction pressure to the left all equates to zero. This is just the border between when over there is acceleration or is not and also we space using the maximum static friction pressure here. Therefore the x-component of the used force is the pressure multiplied by cos Θ due to the fact that we room finding the adjacent leg that this appropriate triangle now and also multiplying cos Θ by the hypotenuse to get the x-component. And also then the maximum static friction force is same to the coefficient of static friction multiply by the common force and so us make substitution"s for each of these terms utilizing what we have written here and we"ll contact this equation number 2. So we have actually Fcos Θ minus μ sF N amounts to 0. Therefore we have two equations and also there room a couple of points that us don"t know yet we can make a substitution indigenous equation 1 and solve it instead for F N and also plug that in up here and also then we"ll be able to figure the end what this used force requirements to be. For this reason here"s equation 1 created again yet version b because we rearranged that a bit: us have included mg to both political parties and included Fsin Θ to both sides and then we finish up with normal force is mg to add Fsin Θ. And also so then us rewrite equation 2, we"ll speak to it variation b whereby we have made a substitution for the normal force and written mg add to Fsin Θ in that is place. For this reason we have actually Fcos Θ minus μ s times mg plus Fsin Θ amounts to 0 then distribute the coefficient of static friction into the bracket and then we end up v this line and also we are fixing for F remember so let"s collection the 2 terms the contain a variable F together and also then aspect out the F for this reason we have F times cos Θ minus μ ssin Θ and also then take the various other term to the appropriate side so we"ll add μ smg to both sides. Okay and then we can divide both political parties by this bracket and we"ll fix for F. Therefore we have actually F is μ smg split by cos Θ minus μ ssin Θ. For this reason that"s 0.1—which is the coefficient of revolution friction of ice cream on ice— time 45.0 kilograms—mass the the block— time 9.80 meter per 2nd squared split by cos of 25 degrees minus 0.1 times sin 25 i m sorry is 50 newtons. Climate in component (b), we are asked well, provided that this applied force continues to be the very same after the starts moving, what will its acceleration be once it"s moving? for this reason we space going to look at equation 2 again; here"s equation 2: Fcos Θ minus μF N amounts to 0 however the μ is gonna be substituted with the coefficient of kinetic friction now because the block is sliding and it"s going come be accelerating so us don"t have a 0 right here anymore, we have an ma here now—this is Newton"s 2nd law— and also our job is to resolve for a therefore we divide both sides by m and we have actually the acceleration is Fcos Θ minus μ KF N over m however this normal force, we require to eliminate that through substituting it v this expression— mg plus Fsin Θ— and so us make the substitution here in ar of F N and also then in ~ this point, you have the right to plug in numbers I made decision to make it a bit an ext clean looking by factoring the end the F between this term and this one after multiply by the negative μ K so that"s why there"s a minus there due to the fact that it"s being multiplied through this minus μ K.

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For this reason we have actually F times cos Θ minus μ Ksin Θ minus μ Kmg everywhere m. For this reason that"s 51.039 newtons— making use of the unrounded prize from part (a); component (a) having only one significant figure due to the fact that this coefficient of static friction has only one significant figure that"s why I created 50 newtons however the unrounded price is 51.039 newtons times cos 25 levels minus 0.03— coefficient that kinetic friction— times sin 25 minus 0.03 time 45.0 kilograms time 9.80 meters per secs squared split by the mass offers us 0.7 meter per second squared.