# Determining displacement

• kara123

#### kara123

Homework Statement
Two students sit in a sled and slide down a hill which gives them an initial horizontal velocity of 9.2 m/s. Their combined mass is 120 kg and the coefficient of friction between the sled and the snow is 0.23. How far will they slide horizontally before coming to a stop?
Relevant Equations
Ek=mv2/2
a=fnet/m
vf2=vi2+2aavd
-began by finding Fnet=Fg-Ff
=1176-270.48
=905.52
acceleration = 905.52/120 kg
=7.546 m/s2

sub numbers into vf2=vi2+2aavd
so clearly it,s wrong because displacement should not be negative in this case

-began by finding Fnet=Fg-Ff
=1176-270.48
=905.52
Forces are vectors, right?

In what direction does gravity act?
In what direction does friction act?
In what direction is the sled moving?
displacement should not be negative in this case
So you get a positive number for force, the equations say that the sled is accelerating forward. You solve them and find that the time that the sled was stationary (according to the equation) was in the past and the corresponding position (according to the equation) was behind where the sled is now. Your algebra was likely correct (I have not checked). Your calculation of the net force and its direction was imperfect.

Kudos for realizing that the displacement should be positive. Sanity checks are a crucial part of problem solving.

Last edited:
Homework Statement:: Two students sit in a sled and slide down a hill which gives them an initial horizontal velocity of 9.2 m/s. Their combined mass is 120 kg and the coefficient of friction between the sled and the snow is 0.23. How far will they slide horizontally before coming to a stop?
Relevant Equations:: Ek=mv2/2
a=fnet/m
vf2=vi2+2aavd

-began by finding Fnet=Fg-Ff
=1176-270.48
=905.52
acceleration = 905.52/120 kg
=7.546 m/s2

sub numbers into vf2=vi2+2aavd
so clearly it,s wrong because displacement should not be negative in this case
Did you draw a FBD of the sled? You should be able to sort out a negative displacement yourself - that just means you are mixing up positive and negative directions somewhere. A diagram should help with that too.

Do you think mass is important? If the mass were ##100kg## would the answer be any different?

Did you draw a FBD of the sled? You should be able to sort out a negative displacement yourself - that just means you are mixing up positive and negative directions somewhere. A diagram should help with that too.

Do you think mass is important? If the mass were ##100kg## would the answer be any different?
well to find the force of friction you have to use the mass so yes if the mass were different the answer would be different

• PeroK
Forces are vectors, right?

In what direction does gravity act?
In what direction does friction act?
In what direction is the sled moving?

So you get a positive number for force, the equations say that the sled is accelerating forward. You solve them and find that the time that the sled was stationary (according to the equation) was in the past and the corresponding position (according to the equation) was behind where the sled is now. Your algebra was likely correct (I have not checked). Your calculation of the net force and its direction was imperfect.

Kudos for realizing that the displacement should be positive. Sanity checks are a crucial part of problem solving.
okay so the only force acting on the sled horizontally is the force of friction so to find the acceleration it would just be -270.48N/120 kg which would equal -2.5m/s2 once subbing that into the equation vf2=vi2+2aavd to solve for displacement I would get an answer of 18.8 so 19m

well to find the force of friction you have to use the mass so yes if the mass were different the answer would be different
The motion is independent of the mass in problems like this. But, we can leave that to one side for now.

• kara123
well to find the force of friction you have to use the mass so yes if the mass were different the answer would be different
okay nevermind i see what your saying since finding the acceleration your dividing by the mass anyways so using any different mass will still provide the same acceleration

okay so the only force acting on the sled horizontally is the force of friction so to find the acceleration it would just be -270.48N/120 kg which would equal -2.5m/s2 once subbing that into the equation vf2=vi2+2aavd to solve for displacement I would get an answer of 18.8 so 19m
That's nearly right. I get something a bit less.

i meant to put the acceleration as 2.25 m/s2 but my displacament is still 18.8m

i meant to put the acceleration as 2.25 m/s2 but my displacament is still 18.8m
What are you using for ##g##?

9.8 N/kg

9.8 N/kg

• kara123
Thanks for all the help!

• PeroK
okay nevermind i see what your saying since finding the acceleration your dividing by the mass anyways so using any different mass will still provide the same acceleration
I know people study physics for different reasons, but to me that is what physics is about! I personally have no interest in whether the sled goes ##5m, 10m## or ##20m##. But, the idea that the motion is independent of the mass is something that does interest me. That's something I do want to understand!

This may or may not be interesting. If I were teaching physics, I would look for problems that to me make sense. In this case, how do we know that the coefficient of friction is ##0.23##? That's seems artificial to me. I would change the problem, so that you are given the distance and asked to calculate the coefficient of friction. That seems to me like something you could actually do in an experiment.

Then, I might replace the initial velocity with an initial height down an incline. Then you could calculate the speed at which the sled leaves the incline as well.

But, I don't teach physics, so I guess you have to focus on the problems that get given to you!

• phinds