A large box of mass M is moving on a horizontal surface at speed v₀. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μₛ and μₖ, respectively. Find an expression for the shortest distance dₘᵢₙ in which the large box can stop without the small box slipping.

Video transcript

Hi, everyone in this practice problem. We're being asked to determine the minimum distance that a cart can cover before stopping with the bag not moving from its position located on top of the cart. We will have a cart moving on a level surface with a speed of 10 m per second, carrying a bag of mass 20 kg. The coefficient of friction between the back and the cart R us equals 0.35 and mu K equals 0.22. The speed of the car goes on decreasing and ultimately, it stops and we're being asked to determine the minimum distance that the car can cover before stopping with the back, not moving from its position. The options given R A 10.6 m B 14.6 m C 3.43 m and D 13.43 m. So we'll model the car as particle. And in this case, let's start with the free body diagram of the car itself. And I'm just going to indicate or denote the car as the circle right here. So the first um force that is going to be acting on the card is going to be the gravitational force or just the weight of the um card or the back itself. And in this case, we will have the gravitational force as FG, which is just essentially M multiplied by G. Next, we will have the normal force acting in counter of the gravitational force. So we have that as N or we denote that as N and then lastly, we will have the frictional force which is going to be horizontally. And I'm going to indicate that with this F right here and it can be FS or FK or static or kinetic friction force. So in this case, let's start with actually using Newton's law to um look into the horizontal and also the vertical axis of this free body diagram. So let's look at the uh vertical axis first, we know that Sigma Fy in the vertical axis will have no acceleration. So in this case, Sigma Fy will equal to zero. And in the Fy di direct or in the Y axis, we have N minus M multiplied by G or FG equals to zero. So in this case, the normal force is just M multiplied by G or essentially just the gravitational acceleration. And then looking into the X direction or Sigma FX, we will know that the uh in the X direction or in the X axis, the card is going to be moving. So Sigma FX is not going to be zero, but instead going to be M multiplied by A X. In this case, um The only force acting upon the X direction is just dis friction force. And I'm just gonna indicate that with FS just like, so which will be equal to M multiplied by A X. Before moving into the next steps. We want to notice that in the practice, in the problem statement, it is given that the speed of the car goes on decreasing and ultimately comes to a stop, which is going to be caused by this friction force F right here. So because it is going against the movement of the car, you want to add a negative sign in front of the FS. Next, the maximum value of the static friction or Fs Max, we want to check FX, FS Max will be equals to us multiplied by N. In this case then because we know that the normal force is M equals to G, then Fs Max will be equals to us multiplied by M multiplied by G which uh after we actually substitute every single um information here, we know that the US is 0.35. But we're going to just keep the M as it is and then the G and substituting it with 9.81 m per second squared. And we're going to substitute the F aspects into our equation from the uh Sigma FX Newton's law and substitute that into the equation that we have there to then obtain negative 0.35 multiplied by M multiplied by 9.81 m per second squared equals to M multiplied by A X. We can cross out the MS and then calculating this, we will then get our um acceleration in the X direction didn't be equal to negative 3. m per second squared. So now that we have obtained the acceleration in the X direction, we will use the kinematic equation which is going to be P one squared equals to P O squared um plus to A X multiplied by delta X. So what we are interested to find or what is being asked in the problem statement is the minimum distance that the card can cover. So what we are interested to find is the delta X. So we wanna rearrange our equation. So the delta X will be equals to P one squared minus V zero squared divided by two um A X. So let's substitute our information. We know in this case, P one is the final velocity which will be 0 m per second because in the problem statement, it is given that the card is going to stop. P nought is going to be equal to 10 m per second, which is given in the problem statement. And the A X is what we have obtained previously. So P one squared is zero squared or 0 m per second squared. Um P knot is 10 m per second squared divide that with two multiplied by negative 3.43 m per second squared. And then calculating this, we will obtain our delta X value to then be equal to 14.58 m or rounding it to one decimal value that will be equal to 14. m which will be the answer to this particular practice problem. So the minimum distance that the car can cover before stopping with the back not moving from its position is 14.6 m which will correspond to option B in our answer choices. So option B will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this video and thank you so much for watching. Bye bye.