Questions Regarding a Cat Going Up a Ramp

[mazia]

Homework Statement

The Pearson question is:

Your cat Goldie (mass 7.40kg) is trying to make it to the top of a frictionless ramp 4.860m long and inclined 33.0∘ above the horizontal. Since Goldie can’t get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 37.50N force parallel to the ramp (there is a set of stairs alongside the frictionless ramp on which you can walk). If Goldie is moving at 1.700 m/s at the bottom of the ramp (assume she has gotten a running start), what is her speed when she reaches the top of the incline?

Relevant Equations

First tried using
F-mg*sin(theta)=ma to solve for a, then using vf^2=vo^2+2as, but then I realized I didn't know what the displacement was.

Then, since I'm asked to find Fnet, I tried
Fn = mg*cos(theta) -> (72.59N)cos(33.0) = 60.88N
Fnet = Fg+Fn+Fapplied -> (72.59N)+(60.88N)+(37.5N) = 171.28N
Fx,net = Fnet*cos(theta) -> 171.28*cos(33.0) = 143.65N but this was incorrect.

EDIT: I've finally found the solution, so here's what I did.
First, calculate Work using the equation: (F-mgsin(theta))*displacement, where F=the force being applied by the push, theta is the angle of the ramp, and displacement is the length of the ramp.
Now that you have the value for work, set it equal to the work-energy theorem:
W = 1/2(mv_f^2)-1/2(mv_o^2). Plug in the values for work, mass and initial velocity, and solve for the final velocity, which is the answer to the question.

I've tried a few different ways but each was wrong. At first, I tried solving for acceleration and then using that to solve for final velocity but that was incorrect. I checked the hints given to me by Pearson which tell me to find the total work first, and looking at the hints for that, they said to find Fx,net which I thought I did correctly but apparently not. Attached is a simplified version of my thought process (including my free body diagram).

I know there's some relation between the force of gravity pointing down (being negative) so I also tried subtracting Fn-Fg and then adding the other forces and finding the component but that didn't work either. I also tried finding the x-component of each force and adding them together, but that didn't work.

Once I have that, I also need to find initial kinetic energy and total work to find the final speed, which I'm not confident in knowing how to do, but I'll cross that bridge when I get there.

 

[PeroK]

If the question is to find a speed, how can the answer be in newtons?

 

[kuruman]

$\dots$ but then I realized I didn't know what the displacement was.

Isn't the displacement 4.860 m as given in the statement of the problem?
Your life would much simpler if you applied the work-energy theorem right from the start.

 

[mazia]

If the question is to find a speed, how can the answer be in newtons?

The speed is the goal, I'm working through the hints to get to that because that's what's being suggested by the homework. I've attached a picture of the homework and the way it's set up.

 

[mazia]

Isn't the displacement 4.860 m as given in the statement of the problem?
Your life would much simpler if you applied the work-energy theorem right from the start.

That's what I'm trying to do. I've attached a picture of the homework, since I forgot to do so, but I'm working through the hints to get to that.

 

[PeroK]

The speed is the goal, I'm working through the hints to get to that because that's what's being suggested by the homework. I've attached a picture of the homework and the way it's set up.

Try the question with the angle of zero and see what happens.

More importantly, a force is a vector quantity!

 

[mazia]

Try the question with the angle of zero and see what happens.

Sorry, but I'm a little confused as to what you mean/why I should do this. So, if I'm using an angle of zero, that makes Fnormal just mg, so the Fnet would be (mg) + Fg + Fapp, and then using the work-energy theorem, I would use that Fnet * displacement (which as another user pointed out is 4.860m) * cos(0) which is just 1, so if theta is 0:
Fnet = Fn + Fg + Fapp -> 72.59+72.59+37.5 = 182.68N
-> * 4.86 = 887.83J
Am I at least partially correct?

 

[kuruman]

Fnet would be (mg) + Fg + Fapp,

It wouldn't. Some of the forces are horizontal and some are vertical. You just don't add them like that. As @PeroK remarked, a force is a vector quantity. How do you add vectors?

 

[PeroK]

Sorry, but I'm a little confused as to what you mean/why I should do this. So, if I'm using an angle of zero, that makes Fnormal just mg, so the Fnet would be (mg) + Fg + Fapp, and then using the work-energy theorem, I would use that Fnet * displacement (which as another user pointed out is 4.860m) * cos(0) which is just 1, so if theta is 0:
Fnet = Fn + Fg + Fapp -> 72.59+72.59+37.5 = 182.68N
-> * 4.86 = 887.83J
Am I at least partially correct?

Using an angle of zero was supposed to highlight the problem with your approach. If the angle is zero, then gravity and the normal force should cancel out, shouldn't they?

 

[mazia]

Using an angle of zero was supposed to highlight the problem with your approach. If the angle is zero, then gravity and the normal force should cancel out, shouldn't they?

Oh, I see, thank you. I wasn't calculating Fg correctly. Or Fn, for that matter. I'll approach the problem again knowing this and update the thread with the results. Thank you so much!

 

[mazia]

Oh, I see, thank you. I wasn't calculating Fg correctly. Or Fn, for that matter. I'll approach the problem again knowing this and update the thread with the results. Thank you so much!

Still got the answer wrong, but here's what I tried.
Fg = mg*sin(theta) -> (7.4*9.81)*sin(33.0) -> 39.54N = Fg
Fn = Fg*cos(theta) -> (39.54)*cos(33.0) -> 33.16N = Fn
Fapplied = 37.5N
Fnet = sum of all forces -> Fg+Fn+Fapplied+Ff -> 39.54+33.16+37.5+0 = 104.15
W = (F*d)cos(theta) -> (104.15*4.860)*cos(33.0) = 424.51 but this is incorrect.

 

[mazia]

It wouldn't. Some of the forces are horizontal and some are vertical. You just don't add them like that. As @PeroK remarked, a force is a vector quantity. How do you add vectors?

Oh I gotcha. So it would be Fapp + (Fn-Fg)?

 

[PeroK]

Oh, I see, thank you. I wasn't calculating Fg correctly. Or Fn, for that matter. I'll approach the problem again knowing this and update the thread with the results. Thank you so much!

To be more precise, you need to express each force in terms of its relevant components. In this case, the components parallel to the slope (which is the direction of motion) are the important ones. The components normal to the slope must cancel as there is no motion in that direction.

 

[mazia]

To be more precise, you need to express each force in terms of its relevant components. In this case, the components parallel to the slope (which is the direction of motion) are the important ones. The components normal to the slope must cancel as there is no motion in that direction.

Uhh..

So Fy,net has to be zero?

 

[PeroK]

Uhh..

So Fy,net has to be zero?

It depends how you define the $y$ direction.

 

[mazia]

It depends how you define the $y$ direction.

The direction perpendicular to the slope.

Anyway, I tried (Fn-Fg) + Fapplied to get the Fnet but it was still wrong so I'm not sure what else to do. Sorry to keep bothering you! This question just really has thrown me through a loop.

 

[PeroK]

The direction parallel to the slope.

Anyway, I tried (Fn-Fg) + Fapplied to get the Fnet but it was still wrong so I'm not sure what else to do. Sorry to keep bothering you! This question just really has thrown me through a loop.

In your first post you calculated $F_n = mg\cos \theta$. That's the normal reaction force, which cancels out the normal force of gravity $- mg\cos \theta$ on the object. Here we have taken positive to be away from the slope.

The important forces, however, are those parallel to the slope. That includes a component of the gravitational force.

 

[mazia]

In your first post you calculated $F_n = mg\cos \theta$. That's the normal reaction force, which cancels out the normal force of gravity $- mg\cos \theta$ on the object. Here we have taken positive to be away from the slope.

The important forces, however, are those parallel to the slope. That includes a component of the gravitational force.

Soo... I'm only worried about Fx,gravity and Fapplied? Add those up and I get my Fnet?

 

[PeroK]

Soo... I'm only worried about Fx,gravity and Fapplied? Add those up and I get my Fnet?

Yes.

 

[mazia]

Yes.

Thank you. So, I took that Fnet, plugged it into the work formula (F*d)cos(theta) -> (70.6N*4.860)cos(33.0) but I still got the answer wrong. I don't understand why I'm struggling so much with this. Thank you for the help so far, though.

This is literally just the hint - I'm not even addressing the actual question yet, I'm just trying to answer the hint. Ugh.
:/

 

[erobz]

Thank you. So, I took that Fnet, plugged it into the work formula (F*d)cos(theta) -> (70.6N*4.860)cos(33.0) but I still got the answer wrong. I don't understand why I'm struggling so much with this. Thank you for the help so far, though.

This is literally just the hint - I'm not even addressing the actual question yet, I'm just trying to answer the hint. Ugh.
:/

The angle between the force and its displacement is not 33 degrees.

 

[PeroK]

Thank you. So, I took that Fnet

How did you calculate $F_{net}$?

, plugged it into the work formula (F*d)cos(theta) -> (70.6N*4.860)cos(33.0)

You need to understand that formula. In this case, it isn't the right formula.

but I still got the answer wrong. I don't understand why I'm struggling so much with this.

I'm not sure where and how you are learning physics, but the key point about forces being vectors seems to have been lost.

 

[mazia]

The angle between the force and its displacement is not 33 degrees.

Ohh. So, drawing a triangle with 37.5N being the force and 4.860m being the displacement, I got arccos(4.860/37.5)=82.56degrees... Is that the right train of thought?

 

[kuruman]

Although it is a very important skill to learn how to deal with components of vectors, this particular problem is on the work energy theorem and does not require you to add forces as vectors. In fact the learning goal of this problem is to "Practice Problem-Solving Strategy 6.1 Work and Kinetic Energy."

Here are the ingredients that you need to put together.

• The work theorem says that the change in kinetic energy is equal to sum of the works done by all the forces acting on an object:$$\Delta K=W_1+W_2+\dots+W_N.$$
• The work done by force $\vec F$ on an object as it is moved by displacement $\vec d$ is $W_F=F~d~\cos\theta.$ Here $F$ is the magnitude of the force vector, $d$ is the magnitude of the displacement vector and $\theta$ is the angle between the force that is doing the work and the displacement vector. Note that the magnitudes $F$ and $d$ are always positive. However, when the angle $\theta$ is less than 90° the cosine is positive; when the angle is 90° the cosine is zero; when the angle is greater than 90° the cosine is negative. So the algebraic sign of the work depends on the angle.

The steps needed to put this together are

1. Draw a diagram showing all the forces acting on the cat.
2. Add to the diagram the displacement vector which is up the incline.
3. Figure out the angle that each force makes with the displacement vector.
4. Figure out the work done by each force.
5. Add all the works to get the cat's change in kinetic energy.

 

[mazia]

How did you calculate $F_{net}$?

You need to understand that formula. In this case, it isn't the right formula.

I'm not sure where and how you are learning physics, but the key point about forces being vectors seems to have been lost.

I calculated Fnet like we mentioned above, Fx,grav + Fapplied.

I'm learning physics at my university and we're emphasizing on forces and vectors, problem is the homework is through Pearson which does not relate to the lectures/current structure of the course, and I also just suck at physics, so I struggle a lot with the homework.

 

[PeroK]

Your other problem, as I see it, is that you don't relate the physics to reality. The applied force is up the slope and gravity acts down the slope. These forces are in opposite directions, yet you add them together. If we took the applied force to be small, then you would still plug the numbers into the equation and get some speed at the top of the ramp. I.e. you would have gravity accelerating the cat up the ramp.

This physics is not an abstract exercise. It's supposed to model reality. I'm not keen on nonsensical problems like this (pushing a cat up a slope?). If you replace the cat with a box being pushed up a ramp you have something fairly realistic.

You need to develop a physical understanding of this and avoid mistakes like having gravity accelerate an object up a slope with increasing speed.

 

[erobz]

Ohh. So, drawing a triangle with 37.5N being the force and 4.860m being the displacement, I got arccos(4.860/37.5)=82.56degrees... Is that the right train of thought?

No. That is not it at all.

 

[PeroK]

Although it is a very important skill to learn how to deal with components of vectors, this particular problem is on the work energy theorem and does not require you to add forces as vectors. In fact the learning goal of this problem is to "Practice Problem-Solving Strategy 6.1 Work and Kinetic Energy."

Here are the ingredients that you need to put together.

• The work theorem says that the change in kinetic energy is equal to sum of the works done by all the forces acting on an object:$$\Delta K=W_1+W_2+\dots+W_N.$$
• The work done by force $\vec F$ on an object as it is moved by displacement $\vec d$ is $W_F=F~d~\cos\theta.$ Here $F$ is the magnitude of the force vector, $d$ is the magnitude of the displacement vector and $\theta$ is the angle between the force that is doing the work and the displacement vector. Note that the magnitudes $F$ and $d$ are always positive. However, when the angle $\theta$ is less than 90° the cosine is positive; when the angle is 90° the cosine is zero; when the angle is greater than 90° the cosine is negative. So the algebraic sign of the work depends on the angle.

The steps needed to put this together are

1. Draw a diagram showing all the forces acting on the cat.
2. Add to the diagram the displacement vector which is up the incline.
3. Figure out the angle that each force makes with the displacement vector.
4. Figure out the work done by each force.
5. Add all the works to get the cat's change in kinetic energy.

Okay, but using acceleration is a valid alternative to work-energy. And, given the OP's problems, it would make sense to get a solution using acceleration to get a grip on the method.

I don't see it makes sense to completely change tack at this stage.

 

[mazia]

Your other problem, as I see it, is that you don't relate the physics to reality. The applied force is up the slope and gravity acts down the slope. These forces are in opposite directions, yet you add them together. If we took the applied force to be small, then you would still plug the numbers into the equation and get some speed at the top of the ramp. I.e. you would have gravity accelerating the cat up the ramp.

This physics is not an abstract exercise. It's supposed to model reality. I'm not keen on nonsensical problems like this (pushing a cat up a slope?). If you replace the cat with a box being pushed up a ramp you have something fairly realistic.

You need to develop a physical understanding of this and avoid mistakes like having gravity accelerate an object up a slope with increasing speed.

I guess so. I'm a pure math person so I'm very used to just seeing numbers and doing stuff with them and not relating them to real world application. For instance, you mentioned the forces being in opposite direction, but adding them together being one of my mistakes - when I hear the words "Sum of Forces", I think "Force + Force + Force + Force...." without first recognizing them as vectors. It's something I'm trying to get the hang of, as this is my first time really studying physics.

 

[PeroK]

I guess so. I'm a pure math person

So am I, so that's no excuse!

 

[mazia]

No. That is not it at all.

I guess I don't know how to do that then. :( I thought I could get there by using trig and drawing the two things as a triangle.

 

[mazia]

So am I, so that's no excuse!

Right but you also have a lot more experience doing this than me so it comes more naturally to you, and things make more sense.

 

[PeroK]

Right but you also have a lot more experience doing this than me so it comes more naturally to you, and things make more sense.

Okay, but back to the problem ...

 

[erobz]

I guess I don't know how to do that then. :( I thought I could get there by using trig and drawing the two things as a triangle.

See post 24 for an explanation of that. There is no sense in me being a parrot.

Also, probably shouldn't jump track to work-energy just yet as @PeroK notes, and get back to Newtons Second Law, and kinematics. It seems like you are having trouble there.

 

[kuruman]

Okay, but using acceleration is a valid alternative to work-energy. And, given the OP's problems, it would make sense to get a solution using acceleration to get a grip on the method.

I don't see it makes sense to completely change tack at this stage.

I completely agree. I have already expressed my thoughts on the importance of being able to add vectors correctly. My intention was not to redirect the path already taken.

I will stand aside until OP has gone down the already established path and obtained the correct answer. After that, I think OP should also obtain the same answer using work-energy considerations in which case I will butt in if necessary.