A wet bar of soap slides down a ramp 9.0 m long inclined at $8.0 ^\circ$. How long does it take to reach the bottom? Assume $\mu_k = 0.060$.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. To figure out how long it takes the soap bar to reach the bottom of the ramp, we need to know what its acceleration will be. And to find its acceleration we'll have to know some things about the forces that are on it. So, there's a kinetic friction force acting against the direction of its velocity. So, it's velocity is down the ramp and so that's why the kinetic friction force is up the ramp. And then there's the component of gravity along the ramp and that's pulling it down. And we have a coordinate system so that we have positive x direction along the ramp down it and then positive y is perpendicular to the ramp. So, in the x direction we have x component of gravity in the positive direction minus the friction force in the negative direction, and that is the net force on the x axis, and so, it equals mass times acceleration. And the x component of gravity is the opposite leg of this gravity triangle. And so, we'll use sine of theta multiplied by the force of gravity, mg, to find this opposite leg. And we know this is theta, by the way, because this angle here, we're going to consider this triangle here, and this angle up in this corner is 90 minus theta because there's a 90 degree angle here. And if this is 90 minus theta and this angle here is 90 as well from the dotted line to the ramp. Then that makes this angle here 90 minus 90 minus theta. So, it's the full 90 minus this thing here which is 90 minus theta which is 90 minus 90 plus theta which is just theta. Alright. What else? Let's get rid of that stuff here but not all of it. So, we know another expression for friction force is coefficient of kinetic friction times the normal force and the normal force will be equal to the component of gravity perpendicular to the ramp, the Fgy. And that's the adjacent leg of this gravity triangle and so, that means we use cosine theta times mg to find Fgy. So, friction force is muK mg cos theta. So, now we're going to substitute into this formula now. Using these two expressions. So, we have Fgx is mg sine theta. And we have that force of friction is muK mg cos theta, and all that equals ma. And we'll divide both sides by m, so, the m cancels everywhere and we put the a on the left hand side and we have it equal to g times sine theta minus muK cos theta. So, it's 9.8 meters per second squared times sine of 8 minus 0.06 times cosine of 8 which gives 0.7816 meters per second squared. And now we turn to kinematics because we've finally figured out what the acceleration of the soap bar will be. And so its displacement will be 1/2 aT squared we usually have V initial times time plus 1/2 aT squared, but the initial speed is zero, so, we can just write this, and we'll solve for T by multiplying both sides by two and dividing by a. And then take the square root of both sides. And T is square root of 2 times displacement over acceleration. So, square root of 2 times 90 meters divided by 0.7816 meters per second squared which is 48.Seconds for the soap bar to reach the bottom of the ramp.