A 1500 kg car skids to a halt on a wet road where mu(k) = 0.50. H...

Question

A 1500 kg car skids to a halt on a wet road where mu(k) = 0.50. How fast was the car traveling if it leaves 65-m-long skid marks?

Accepted Answer

Hi, everyone in this practice problem. We are being asked to find the speed of a cable reel where we'll have a truck carrying drum shaped cable reel of electrical wires losing control and banks on the roadside wall. One of the cable reels rolls down from the truck and stops m from the truck. We're being asked to find the speed of the cable reel if its mass is 550 kg and the coefficient of kinetic friction is 0.45. The options given for the speed of the cable reels are a 4. m per second. B 8.82 m per second. C 16.3 m per second and D 31. m per second. So we will consider the cable reel as a particle. And therefore the pictorial representation or the free body diagram will then come out to B. So the cable real, I'm just gonna represent it as the sphere or circle right here. First, it will have the gravitational four FG just like. So which is essentially just the mass of the cable real multiplied by the gravitational acceleration. It will also have the normal force acting in reverse or countering the gravity with this N right here. And lastly, it will have the frictional force which is going to be represented by this FS just like. So, so um the X and Y or axis is just a normal vertical and horizontal. And here the kinetic friction is going to be responsible for providing the negative acceleration due to which the reel will come to a halt. So we wanna apply Newton's 1st and 2nd law at first in the X axis and then in the Y axis as well. So let's look into the Y axis because it's a little simpler. We know that uh we will assume that it is not going to move in the Y axis, the cable reel. So in the Y axis, Sigma Fy is going to be equals to zero. So zero will be equals to N minus FG and substituting FG, we will get N minus M multiplied by G or N will essentially be equal to M multiplied by G just like. So, so let's hold on to this for a little while. And then we wanna move or look into Sigma FX uh or in the X direction where Sigma FX will actually be equal to M multiplied by A X because the cable role is going to be moving in the X direction. So in the X direction, the only force that is going to be um acting on this thing is just uh the friction. So in this thing, friction is going to be FK. So FK will be equal to M multiplied by A X just like. So, so FK can be obtained by Mu K multiplied by N which will equals to M multiplied by A X. So let's substitute our N value that we just obtained or an expression that we just obtained from the Sigma Fy um equation. So let's do it right that so mu K will then be equal to or mu K multiplied by M multiplied by G will be equal to M multiplied by A X. So in this case, we will then be able to cross out the MS. So A X will then be equal to mu K mu K multiplied by G just like. So, so let's substitute the UK and G value. The kinetic friction or coefficient of kinetic friction is given to be 0. multiplied by G which is 9.81 m per second squared. Calculating this, the acceleration will then be equal to 4. m per second squared. So we know that the acceleration is what is responsible for the um cable role to come to a halt or a stop. So in this case, the actual value for A X will then be equal to negative 4.41 m per second squared because it is essentially is countering the movement of the actual cable reel itself. So let's now apply the kinematic equation where VF squared equals to V I squared plus two A X multiplied by delta X. And in this case, what we want to find is the um initial V I or the initial velocity. So let's actually uh substitute all of our values. So we know at VF will be equals to 0 m per second because the cable reel comes to a halt. And then the A X is negative 4.41 m per second squared. And then the delta X is given to be 30 m from the problem statement. So let's substitute all of that values. So V I squared will then be equal to zero minus two, multiplied by A X is going to be negative 4. m per second squared multiplied by delta X which is going to be 30 m. So calculating this V I uh squared will then be equal to 264.6 m squared per second squared. And then taking the square root of that for V I. So V I will be equal to the square root of 264.6 m squared per second squared, which will come out to A V I value of 16.3 m per second. So that will be the final uh answer that we have, which is the V I or the initial velocity for the cable reel being equal to 16.3 m per second, which will correspond to option C in our answer choices. So option C 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 one And thank you so much for watching. Bye bye.

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A 1500 kg car skids to a halt on a wet road where mu(k) = 0.50. How fast was the car traveling if it leaves 65-m-long skid marks?

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