Calculate the coefficient of friction of a box sliding down a hill?

AI Thread Summary
A block sliding down a 20° slope with an initial speed of 6 m/s comes to a stop after traveling 11 m, prompting a calculation of the coefficient of kinetic friction. The discussion emphasizes using conservation of energy, where the initial energy is a combination of potential and kinetic energy, due to the presence of friction. The equation derived for the coefficient of friction incorporates both gravitational forces and the slope's angle. Participants clarify that mechanical energy is not conserved in this scenario because of friction, leading to confusion about energy states. Ultimately, the correct approach involves recognizing both kinetic and potential energy throughout the block's motion.
nerdgirl14
Messages
6
Reaction score
0

Homework Statement



A block slides down a hill. The hill has a gentle 20° slope with respect to
the horizontal. The block has an initial speed of 6 m/s. If the block comes to a stop
after traveling a total distance of 11 m along the slope, what is the coefficient of kinetic
friction between the block and the hill?

Homework Equations



F=ma
Fkk*n

The Attempt at a Solution



The first thing I did was draw a FBD. The box is going forward in the positive direction so the kinetic friction is acting in the opposite direction. There is force of gravity acting on the box and normal force, these forces are in the opposite direction on y axis. My x-axis with was with respect to the slop, so kinetic friction is in the negative direction and the box is in position direction, on the x axis.
Now I don't know what to do next...help?
 
Physics news on Phys.org
Use conservation of energy.
 
Okay so use
1/2*m*v2=m*g*h
and I solve for m, this way then I can find gravity and find the two components of gravity. The y component of gravity will give the normal force. But then how do i find the kinetic friction?
 
What is initial energy and what is the final energy of the block. The difference is work done, which is friction.
 
Ok wouldn't the m cancel out with the conservation of energy?
 
Yes, All have m as factor.
KE=\frac {1}{2}mv^2
PE=mgh
friction=μN=μmgCosθ
 
Oh I see now
So I can set up the equation like this:

1/2mv2-mgh=μmgCosθ

Isolating for μ:

μ=(v2/2-gh)/gcosθ


μ=(6m/s2/2-(9.80 m/s2)(11m))/(9.80 m/s2)cos 20°

μ=-9.75

Is this correct?
 
Your equation should be both sides with equal units, joules.
Your initial energy should be PE+KE.
 
How come the initial energy is PE+KE? I thought originally the box was sliding down so all energy is KE and then it comes to a stop so all energy is PE
 
  • #10
azizlwl said:
Use conservation of energy.
Mechanical energy is not conserved in this problem; there is friction.
 
  • #11
nerdgirl14 said:
How come the initial energy is PE+KE? I thought originally the box was sliding down so all energy is KE and then it comes to a stop so all energy is PE
There is always both kinetic and potential energy, even though at some points one or the other may be zero. Of course, by `always' I mean always when there is something around to give rise to potential energy.
 
  • #12
tms said:
Mechanical energy is not conserved in this problem; there is friction.

Work-Energy Principle.
 
  • #13
Never mind.
 
Last edited:
Back
Top