Sledding Down a Hill: Calculating Friction and Velocity

[DeepPatel]

Homework Statement

A man (75kg) is sledding down a 20m high hill that is at an angle of 30 degrees. He is given a push that has him initially going 7m/s. He experiences a force of friction of 22N on his way down the hill.

a) How much work is done by friction on his way down the hill?

b) How fast is he going when he gets to the bottom?

Homework Equations

W=Fdcos(Θ), ME_i + W = ME_f

The Attempt at a Solution

For (part a) I thought you had to use 1/2mv_i² + W = mgh but I feel like I am messing something up with the potential and kinetic energy.

For (part b) I know you solve for final velocity but I'm not sure if you use 1/2mv_i²+ mgh + W = 1/2mv_f² after getting the answer for work.

 

[ramzerimar]

Homework Statement

A man (75kg) is sledding down a 20m high hill that is at an angle of 30 degrees. He is given a push that has him initially going 7m/s. He experiences a force of friction of 22N on his way down the hill.

a) How much work is done by friction on his way down the hill?

b) How fast is he going when he gets to the bottom?

Homework Equations

W=Fdcos(Θ), ME_i + W = ME_f

The Attempt at a Solution

For (part a) I thought you had to use 1/2mv_i² + W = mgh but I feel like I am messing something up with the potential and kinetic energy.

For (part b) I know you solve for final velocity but I'm not sure if you use 1/2mv_i²+ mgh + W = 1/2mv_f² after getting the answer for work.

a) What's the basic definition of work?

b) Have in mind that, when the man is pushed, he is at the top of the hill so he has both kinetic and potential energy. When he gets to the bottom, he has only kinetic energy. It's just conservation of mechanical energy.

 

[haruspex]

For (part a) I thought you had to use 1/2mvi2 + W = mgh

You don't have enough information to tackle it from that end.
What other equations can you think of for work done by a force?

 

[DeepPatel]

a) What's the basic definition of work?

b) Have in mind that, when the man is pushed, he is at the top of the hill so he has both kinetic and potential energy. When he gets to the bottom, he has only kinetic energy. It's just conservation of mechanical energy.

a) Using W=Fdcos(theta) I don't know what the distance is, do I solve for the hypotenuse?

b) If he only has kinetic energy then wouldn't it only be 1/2mv²_i + W = 1/2mv_f

 

[haruspex]

a) Using W=Fdcos(theta) I don't know what the distance is, do I solve for the hypotenuse?

Yes, why wouldn't you?

 

[DeepPatel]

You don't have enough information to tackle it from that end.
What other equations can you think of for work done by a force?

Yes, why wouldn't you?

Did that and used W=Fdcos(theta) and got 190.525J, now I'm just confused on what I am missing for part b.

 

[haruspex]

Did that and used W=Fdcos(theta) and got 190.525J

Looks wrong to me. Please post your working.

For part b), the equation you originally posted was correct, but we need to see how you apply it.

 

[DeepPatel]

Looks wrong to me. Please post your working.

For part b), the equation you originally posted was correct, but we need to see how you apply it.

a) I did it over and first I did cos(30) * 20 to solve for the hypotenuse and got 17 m. Then I did 22 * 17 * cos30 and got 324J.

 

[haruspex]

I did cos(30) * 20 to solve for the hypotenuse

That is not going to give you the hypotenuse. Did you draw a diagram?

 

[DeepPatel]

Yeah I did, the first time I did it I just did the sin(30) = 20/x and got 10 for x.

 

[ramzerimar]

Are you sure that sin(30) = x/20?

 

[DeepPatel]

Sine is opposite over hypotenuse and we are giving the opposite which is 20m unless I am putting my angle in the wrong spot.

Sin(30) is equal to 0.5 which means that x would have to be 40 which I calculated wrong in the first place.

 

[ramzerimar]

Sine is opposite over hypotenuse and we are giving the opposite which is 20m unless I am putting my angle in the wrong spot.

Sin(30) is equal to 0.5 which means that x would have to be 40 which I calculated wrong in the first place.

Yes, x must be 40.

 

[DeepPatel]

Alright so knowing that using W=Fdcos(theta) I should get 762J.

part b) If that is work then I can now solve for v_f by using the original formula and get 427.32 m/s? Seems a bit large to me.

 

[haruspex]

using W=Fdcos(theta) I should get 762J.

Still wrong. How do you get that? What is theta there?

 

[DeepPatel]

Still wrong. How do you get that? What is theta there?

Did 22N*40m*cos(30) and got that.

 

[ramzerimar]

Did 22N*40m*cos(30) and got that.

Regarding item (b), remember to take the square root!

 

[haruspex]

Did 22N*40m*cos(30) and got that.

In the standard equation W=Fd cos(theta), how is the angle theta defined?

 

[DeepPatel]

In the standard equation W=Fd cos(theta), how is the angle theta defined?

Oh I forgot the angle is defined by the direction of the force and distance, in this case the force and distance are opposite to each other so would the angle be 180 degrees?

 

[haruspex]

Oh I forgot the angle is defined by the direction of the force and distance, in this case the force and distance are opposite to each other so would the angle be 180 degrees?

Whether they are opposite or in the same direction depends on how you are defining the positive directions, but yes, that will give the right magnitude.

 

[DeepPatel]

So then work would be equal to 880J for part a.

 

[haruspex]

So then work would be equal to 880J for part a.

Yes.

 

[DeepPatel]

Yes.

Now I plug that into the original question 1/2mv_i²+ mgh + W = 1/2mv_f², solve for final velocity and I got 21.66 m/s

 

[haruspex]

Now I plug that into the original question 1/2mv_i²+ mgh + W = 1/2mv_f², solve for final velocity and I got 21.66 m/s

Did the work done by friction make him go faster?

 

[DeepPatel]

I don't think it should but if the equation I have is correct I'm not sure where I am going wrong.

Since it is asking for the speed at the bottom I do think it should have a higher speed.

 

[haruspex]

I don't think it should but if the equation I have is correct I'm not sure where I am going wrong.

Since it is asking for the speed at the bottom I do think it should have a higher speed.

Let me put it another way... in the equation you used for the man's mechanical energy at the bottom, did the work done by friction make a positive or negative contribution?

Valuable tip: never apply equations blindly and trust the answers. There are too many ways you can accidentally misapply them. Part of your brain has to keep asking "does this make physical sense?"

 

[DeepPatel]

Let me put it another way... in the equation you used for the man's mechanical energy at the bottom, did the work done by friction make a positive or negative contribution?

The work done by friction was negative, yes.

 

[haruspex]

The work done by friction was negative, yes.

Then you should have got a smaller answer. Please post all your working.

 

[ramzerimar]

The work done by friction was negative, yes.

Try this: if there was no friction, what should your final speed be?
So, if you have negative work done by friction, your speed should be smaller than that.

 

[DeepPatel]

Then you should have got a smaller answer. Please post all your working.

Try this: if there was no friction, what would your final speed be?
So, if you have negative work done by friction, your speed should be smaller than that.

I had added 880J and I think I had to subtract. If I fixed the errors I now got an answer of 20.549 m/s

(1/2)(75)(7²) + (75)(10)(20) - 880 = 1/2 (75) (v²_f)

 

[haruspex]

I had added 880J and I think I had to subtract. If I fixed the errors I now got an answer of 20.549 m/s

I get slightly less, but that should be close enough. Maybe you used a larger value for g.

 

[DeepPatel]

I get slightly less, but that should be close enough. Maybe you used a larger value for g.

Yeah our teacher prefers we use 10 m/s² instead of 9.81m/s² to save us time and make it easier.