A block of mass m is at rest at the origin at t = 0. It is pushed with constant force F₀ from 𝓍 = 0 to 𝓍 = L across a horizontal surface whose coefficient of kinetic friction is μₖ = μ₀ ( 1 - 𝓍/L ) . That is, the coefficient of friction decreases from μ₀ at 𝓍 = 0 to zero at 𝓍 = L.

b. Find an expression for the block's speed as it reaches position L.

Question

A block of mass m is at rest at the origin at t = 0. It is pushed with constant force F₀ from 𝓍 = 0 to 𝓍 = L across a horizontal surface whose coefficient of kinetic friction is μₖ = μ₀ ( 1 - 𝓍/L ) . That is, the coefficient of friction decreases from μ₀ at 𝓍 = 0 to zero at 𝓍 = L.

b. Find an expression for the block's speed as it reaches position L.

Accepted Answer

Hi, everyone. In this practice problem, we're being asked to determine an equation for a box's velocity moving from point X equals to L to X equals to two L. We will have a box with a mass M pushed with a constant forced F along a frictionless horizontal surface. The box is initially at rest at the origin or at X equals to zero at FX equals to L. The surface actually becomes rough and the coefficient of kinetic friction is then Mu K equals to mu not multiplied by open parentheses, one minus X divided by L closed parentheses. The force applied to the box is maintained at a constant magnitude of F throughout its whole motion. And we're being asked to determine an equation for the box's velocity moving from X equals to L to X equals to two L. The option for the box's velocity equation are a square root of open parentheses for L multiplied by F divided by M close parenthesis, B square root of open parenthesis to L multiplied by F divided by M close parenthesis, C square root of open parentheses for L multiplied by F divided by M minus mu not multiplied by G close parenthesis and D square root of open parenthesis to L multiplied by F, divided by M minus mu not multiplied by G cross parentheses. So the way we want to tackle this problem is to by applying Newton's law, so we wanna apply Newton's second law in this case to our system, which is going to be the box to get our equation. So first, I'm gonna draw out the system that we have. So first, what we have is the box at point we're gonna focus at point X equals to L and X equals to two L where the surface or the horizontal surface is actually rough just like. So, so the box will actually be pushed from point X equals to L to point X equals to two L with a constant force of F. So I'm gonna write down the force here or note down the force here. So that will be the force F and through out point X equals to L to X equals to two L, there will be a friction force going counter to F, which is going to be represented by F K just like. So, so this is going to be the diagram or the free body diagram of the system that we have. So let's start with actually doing the Newton second law. So according to Newton's second law, in this case, we're looking at the X axis. So at the X or horizontal axis Sigma FX will actually equals to M multiplied by A X where M is the mass of the box and A X is the acceleration in the X direction. So there are two different forces acting on the X direction which is F minus F K, where we have the convention for the force going to the right to be positive. So F minus F K will equals to M multiplied by A X. And then we wanna look at uh Newton's Law as well in the Y direction. So in the Y axis, we have only two different force as as well, which is one is represented by the normal force. And the other one is represented by the weight, which is just M multiplied by G and the box itself is not moving in the Y direction. So Sigma F Y will actually equals to zero. And Sigma F Y is going to be N minus M multiplied by G equals to zero. And that will mean that N will equals to M multiplied by G just like. So, and the normal force here is actually needed for us to actually solve the friction force. Because if we recall friction force can be calculated calculated by multiplying the coefficient of kinetic friction UK multiplied by the normal force. And I'm just gonna keep the equals to M multiplied by A X here. And the normal force can be substituted with our equation that we found from the Y axis Newton's Law and the UK can be substituted from the equation that is given in the problem statement. So I'm going to substitute those two into our equation. So then F minus first is going to be our UK. So UK is mu not multiplied by open parenthesis, one minus X divided by L. And I'm gonna multiply that with M multiplied by G equals to M multiplied by A X just like. So, all right. So this is the equation that we found uh right now. And what we want to do next is to pretty much just define everything in the equation by M to simplify this. So I'm just gonna cross out the M from this two. So, but then the F will then be F over M, I'm going to rewrite that for simplicity. So then we will have F divided by M minus Muno multiplied by open parentheses, one minus X divided by L co parentheses multiplied by A G equals to A X just like. So I'm gonna call this equation one so that we can refer to dis spec again, further down the line. All right. Next, we want to recall that A X actually equals to D V X divided by D T and D V X divided by D T can actually be expanded into D V X over D X multiplied by D X over D T. And if you notice the X over D T actually S equals to V X. So we want to substitute that here so that A X actually equals to V X multiplied by D V X over D X just like. So next, we can actually bring our D X around to A X so that we get A X multiplied by D X equals to V X multiplied by D V X just like. So, so the point of us doing this is to actually get the equation for velocity. This is how we are bringing the velocity into our equation or our expression. So to get the actual velocity now that we are looking at this, we still have the D V X here, we will have to integrate this equation. So we want to integrate this from both sides. So I'm gonna write, integrate. So integrating this, the integral for of A X D X will equals to the integral of V X D X or V X D V X and the A X D X can be, can be obtained from our A X equation or equation one. So I'm gonna uh rewrite equation one into our left side of our equation here. So the integral of F divided by M minus M not multiplied by uh open parentheses, one minus X divided by L plus parenthesis multiplied by G of multiply that by D X will equals to the integral of V X multiplied by D V X, the integral of the life, the, the right side is actually pretty simple. So if you wanna uh recall the principles of integral, the integral of X D X will just be half X squared plus C. So in this case, the integral of V X D V X is just going to be half P X squared plus C just like. So, all right, so we get this equation for the right side, but we have to solve the integral for the left side. And if you are looking at the equation on the left side closely, then you will be able to tell that there will be three different terms that we have to integrate separately. The first one is going to be this F divided by M and then you not multiply it by G multiply it by one. And then lastly, you not multiply by G multiply by X divided by L. So I'm gonna rewrite this so that it's easier for us. So in the girl for F divided by M minus mu not multiplied by G plus mu not multiplied by G multiplied by X divided by L. All of that with respect to D X well equals to half three X squared plus C just like. So, all right. So the integral for a divided by M and multiply it by G is just going to be an integral of a constant. So the integral of a constant C with respect to D X will just equals to C X plus C essentially. And in this case, then we will then have F divided by M X minus mu not multiplied by G X plus. Uh mu not multiplied by G divided by L is actually going to just be a constant. So I'm gonna write that down first. And the X here is going to follow the integral principle that we have used previously. So that will then be a half X squared. And I'm gonna lump all the C on the left side with the CS on the right side and just put it all to one side, which is going to just be on the right side. So this will then equals to half V X squared plus a big C of all the Conant lump together. All right. So this is the equation that we are going to have and we're going to next find what C is. So using our um at rest condition at X equals to zero, we have V X equals to zero. So that means we want to substitute X and V X into this equation to get what C is. So substituting that N, the first term will then be zero minus zero plus zero will equal to zero plus C. So that means our C will then equals to zero. So finally, our equation will then equals to F divided by M X minus mu not multiplied by G. I'm going to lump the mu not multiplied by G together. I'm just gonna put this as a plus so that it's easier for us. So this will then actually be X squared divided by two L minus X just like. So and this will equal to V X squared divided by two. And this is the equation that we have for. Now, what we are interested at is an equation for just V X. So we wanna rearrange this to have an equation for just V X. So V X will then actually be the square root of two multiplied by open parenthesis. FX divided by M plus U knot multiplied by G multiplied by open parentheses. X squared divided by two L minus X close parenthesis. So this will be the equation for P X as a function of X. And what we have to find is to get or to find the box's velocity at X equals to two L. So at X equals to two L, we just want to substitute the two LS into our equation. So V X of two L or at X equals to L A to two. So this is the V X with a function of X. Well, that equals to the square root of two multiplied by two L F divided by M plus U not multiplied by G multiplied by open parenthesis. Uh two L squared will be four L squared divided by two L minus two L. This is just two L if we simplify this. So the right side will actually become zero. So then R V X of two L will actually equals to the square root of two multiplied by two L F divided by M. And that will be B X of two L equals to the square root F four L F divided by M. And that will essentially be the final answer for the velocity of our box at V X equals to two L which is square root of four L F divided by M which will actually correspond to option A in our answer choices. So option A will be the answer to this practice problem with the velocity equation for the boxes moving from X equals to L to X equals to two L equals to the square root of four L F divided by M. So that will be it for this video. If you guys still have any sort of confusion on this particular topic, please make sure to check out our other lesson videos and that will be it for this one. Thank you.

Verified Solution

Key Concepts

Newton's Second Law of Motion

Friction and its Coefficient

Work-Energy Principle