A 5.0 kg wooden sled is launched up a 25° snow-covered slope with an initial speed of 10 m/s.

a. What vertical height does the sled reach above its starting point?

Question

A 5.0 kg wooden sled is launched up a 25° snow-covered slope with an initial speed of 10 m/s.

a. What vertical height does the sled reach above its starting point?

Accepted Answer

Hi, everyone in this practice problem, we're being asked to calculate the vertical height reached by a cart with respect to its starting point where we'll have a steel cart weighing at kg pushed upwards on a 20 degrees inclined steel slope at a speed of 12 m per second. We are being asked to calculate the vertical height reached by the cart with respect to its starting point. And the options given are a 21.4 m B 7.32 m C 3.37 m and D 12 m. So we will use the particle model and consider that the steel car is going to be a particle. I'm going to start us off with um drawing a free body diagram or creating a pictorial representation of our situation in the problem statement. And I'm gonna start with drawing our incline slope. So the incline slope is going to be inclined at a 20 degrees angle. And on the slope, we will have a steel cart weighing 25 kg being pushed upwards which I'm just going to represent with a dot Because essentially we are imagining or we're using the steel card as a particle, I'm going to draw a X and A Y axis so that it's easier for us to actually draw our free body diagrams. And the X and A Y axis is going to actually be alongside the incline of the slope and also perpendicular to the slope. All right. So let's start with uh actually tackling the different forces that is going to be acting upon our steel card. So the first one is going to be the gravitational force which is going to just be the weight of our steel card, which is M multiplied by G or essentially F G. The second one, we will have a normal force acting perpendicular to the incline slope or alongside the direction of our Y axis. This is going to be our end or normal force. And then the last force that is going to be acting upon, this is the friction between the steel car and the steel slope, which is going to be opposite to the direction at which the steel card is pushed on. So it is going to be going to the left side on the X axis just like. So and this will be represented by F R just like so, so 20 degrees is going to be the slope or the incline slope. And I'm going to also write down that this angle right here is also going to actually be degrees because of the fact that our 20 degrees, our right angle on our slope and also the degree next to the 20 degrees that we just calculated is going to sum up to 100 and 80. So their angle right here is actually going to be degrees. But because uh the two axis, the two X and y axis has to meet 90 degree angle, so we can determine that our angle between the y axis and our gravitational force is going to be 20 degrees. So that will be the final uh representation or the final free body diagram that we have. And now let's start actually um considering the motion of the car by actually tackling the Newton Second law. So according to the Newton Second law, Sigma F will equals to M multiplied by a on a certain axis. We will have to look for both axis. In this case, I'm going to start with the X axis or X direction. So Sigma FX will equals to M multiplied by A X and the forces acting on the X axis are going to be first F R and also second the projection of F G. So I'm gonna uh draw the projection of F G here. So F G on the X axis is then going to actually be I'm gonna write it up here in black is then going to be F G multiplied by sine of 20 degrees because it is opposite to the 20 degrees angle or opposite to the node angle. So that will be the F G projection on the X axis. So then our Sigma FX will then be because both F G S uh sine 20 degrees and F R will be going to the left. And the convention is that the forces acting going to the left is going to be negative. So then we'll have negative F G multiplied by a sine of degrees minus F R equals to M multiplied by A X F G is just uh M multiplied by G. So negative M G sine 20 degrees minus F R F R is just mu R multiplied by the normal force equals and multiplied by A X. And this will be our first equation. I'm gonna just represent it as equation one for us to make it easier for us to actually refer down the line. Next, you want to look at the Newton second law in the Y axis. So in the Y direction, Sigma F Y will equals to M multiplied by A Y. But in this case, because we define our Y axis uh perpendicular to our incline slope, there will be no movement in the Y axis. So in this case, M multiplied by A Y is actually going to equals to zero Newton. Uh the two forces acting on F on the Y direction to count for Sigma F Y is going to be first the normal force and also the projection of F G in the Y axis which will then be F G multiplied by coside of 20 degrees. So the normal force is going upwards. So, and minus F G going downwards multiplied by cosine of 20 degrees will then equals to zero. So then R N or normal force will equals to I'm gonna put the arrows there. N will equals to F G multiplied by cosine 20 degrees. And F G is just M multiplied by G. So N will equals M multiplied by G multiplied by cosine of 20 degrees. This will be our second equation. And what we wanna do next is to actually combine equation one and equation two or essentially substitute N into our uh normal force in the first equation here. So that we can then get negative M G sine degrees minus U R multiplied by M G cos sine 20 degrees equals MA X. So this is combining one and two. Awesome. So then we have uh the mass or M in every single term in our equation. So I'm just gonna cross that out real quick. So then our equation will then be negative G sine 20 degrees minus U R G multiplied by cosine of degrees equals A X. We want to actually then substitute all of our known information here because we know what G is and mu R, we will know Mu R because it is just a common fact for the friction coefficient for steel on dry steel, Mu R is going to be 0.2. So then R A X will then be negative G which is negative 9.81 m per second squared multiplied by a sign of 20 degrees minus M R which is 0.2, multiplied by G which is 9.81 m per second squared multiplied by a cosine of 20 degrees. And then that will give us uh acceleration on the X axis to be negative 3.37 m per second squared just like. So, so now we have uh gotten our acceleration. And what we need to do next is to actually calculate our vertical height. So calculating our vertical height, we will employ the kinematic equation. And the equation that we're going to use is going to be V F squared equals to V I squared plus two A X multiplied by D D delta X. We can rearrange this equation so that we have an equation for delta X because that is what we are interested at delta X is there going to be V F squared minus V I squared divided by two A X. So we can substitute all of our node information. We know that by the end of the movement our uh steel cart will be coming to a rest. So in this case, our V F squared is going to be zero and R V I squared or R V I is known to be 12 m per second in the problem statement. So this will then equals to zero minus 12 m per second parentheses squared. All of that divided by two multiplied by a negative 3.37 m per second squared, just like soap. And that will actually give us delta X to be 21.4 m. So this will be the distance traveled along the inclined plane or essentially 21.4 m is going to be this distance right here. So this is going to be delta X which is apparently 21.4 m. And what is being asked is the vertical height. So the vertical height to get the vertical height, we will have to use trigonometry. So H or the vertical height will equals to delta X multiplied by, we want to project delta X so that we get just the height. So in that case, we want to multiply delta X by sine of 20 degrees. So H will then equals to delta X multiplied by sine of 20 degrees, which will then equals to 21.4 m multiplied by sine of 20 degrees which come out to be 7. meter. So the vertical height will actually be just 7.32 m and that will correspond to option B in our answer choices. So option B with the vertical height reached by the card with respect to its starting point being 7.32 m will be the answer to this particular practice problem. So if you guys still have any sort of confusion, please make sure to check out our other listed videos on similar topic and that'll be it for this one. Thank you.

A 5.0 kg wooden sled is launched up a 25° snow-covered slope with an initial speed of 10 m/s. a. What vertical height does the sled reach above its starting point?

Verified Solution

Key Concepts

Kinetic Energy

Potential Energy

Conservation of Energy