A baggage handler throws a 15 kg suitcase along the floor of an a...

Question

A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitcase slides 2.0 m before stopping. Use work and energy to find the suitcase's coefficient of kinetic friction on the floor.

Accepted Answer

Everyone in this problem. A three kg box is pushed across a horizontal surface with a speed of 1.5 m per second. The box travels a distance of 1.25 m before coming to a stop. We're asked, what is the coefficient of kinetic friction between the box and the surface? The answer traces were given are a 0. B 0.09 C 0.06 ND 0.6. Now we're looking for the coefficient of kinetic friction and we know that that comes up in our force of friction. OK. So we have the force. We're given some information about the mass as well. Recall that we can relate the force to the mass and the acceleration through Newton's second law. We have that the sum of the forces is equal to the mass multiplied by the acceleration. So let's start by finding the acceleration. A We know that the initial speed of the box is 1.5 m/s. The box comes to a stop. So the final speed is 0m/s. It travels a distance of 1.25 m before coming to a stop. And so delta X, the displacement is 1.25 m and we wanna find the acceleration A. Now we have three known values so we can use our U AM or kinematic equations to solve for the acceleration. A. We're gonna choose the equation that doesn't have the time t in it because we don't have information about that. And that's not what we're looking for. We have VF 2ared is equal to V not squared plus two, multiplied by a multiplied by delta X substituting in our values, we have zero is equal to 1.5 m per second squared plus two, multiplied by a Multiplied by 1.25 m. We can move our 1.5 m per second square term to the left hand side. We have negative 2.25 m squared per second. Squared Is equal to 2. m multiplied by a. If we divide by 2.5 m, we get that the acceleration a is equal to negative 0.9 m/s squared. Now, we have our acceleration A so we can use Newton's second law. Let's draw out a free body diagram to get an idea of the forces acting on this box. It's on a surface. And so we know we have the normal force pointing upwards. Always we have the force of gravity pointing downwards. We're gonna take the right to be our positive direction. In the direction of motion, which means that the force of friction acts to the left, it opposes the motion and that's gonna be the only force in the direction the box is initially pushed, but it's pushed and then that force no longer acts on it while it's sliding. Ok. So the only force acting on the box while it's moving is that force of friction. So we want the force of friction. So let's look at our sum of forces in the X direction. So we have that the force sum of forces in the X direction is equal to the mass multiplied by the acceleration in the X direction in the X direction, we only have the force of friction. OK. So F K is equal to M multiplied by A X. The force of friction. F K recall is equal to minus M K oops and this should be minus the force of friction because it's acting in the negative direction. OK. So minus F K is equal to the mass multiplied by the acceleration in the X direction. So we have minus UK multiplied by the normal, that's the friction force is equal to mass M multiplied by the acceleration A X. Now we know the mass, we know the acceleration in the X direction. OK. The acceleration we've found using kinematics is in the X direction because that's the only direction where we have motion. All we need now is the normal force. N in order to find mu K, the norm of course acts in the Y direction. So let's look at the sum of forces in the right direction, we have no motion in the Y direction. And so the sum of forces is going to be equal to zero are equilibrium condition. Now, in the Y direction, we have the normal fork pointing upwards, we have the force of gravity pointing downwards. So we have N minus F G is equal to zero. It tells us that the normal force end is equal to the force of gravity F G. And recall that the force of gravity is given by the mass M multiplied by the acceleration due to gravity. G, we can move back to our direction. Again, we're still looking for this coefficient of kinetic friction. UK we have the normal force and now we have minus mu K multiplied by M multiplied by G is equal to M multiplied by A X. We can divide by M on both sides. We have that UK is equal to negative the acceleration which is negative 0.9 m per second squared Divided by acceleration due to gravity 9.8 m/s squared. And we get a coefficient of kinetic friction of 0.092. Comparing this to our answer choices. OK to two decimal places. We found that the coefficient of kinetic friction is approximately 0.9 which corresponds with answer choice B Thanks everyone for watching. I hope this video helped see you in the next one.

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A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitcase slides 2.0 m before stopping. Use work and energy to find the suitcase's coefficient of kinetic friction on the floor.

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