Solving Ramp with Friction Homework Problem

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Homework Statement



When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.15 m/s. The mass stops a distance S2 = 2.5 m along the level part of the slide. The distance S1 = 1.15 m and the angle θ = 30.30°. Calculate the coefficient of kinetic friction for the mass on the surface.

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Homework Equations



1/2mv2 + mgh = ?

The Attempt at a Solution



I do not know where to start..
 
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SammyS said:
PEI + KEI = PEF + KEF - WFriction, where WFriction is the work done by friction.

WFriction is itself negative, so subtracting it from zero will give a positive number.

PEi can be found using mgh, in which h is found using S1 and the angle

KEi is 1/2mv2 in which v is given

PEf is where I have doubt

KEf equals 0 since the object is at rest

Am I correct?
 
SammyS said:
If h is the elevation above the level, then what is h on the level?
So, PEi + KEi = - Wfriction?

How do I calculate work done by friction?

Is it considering just the 1.15m from the ramp, the 2.5m or both?
 
Last edited:
I cannot get an answer. How do I use friction?
 
alfredo24pr said:
I cannot get an answer. How do I use friction?
The normal force for S2 is obviously mg.

What is the normal force for S1, when the block is on the ramp?

I do have a question about the problem. It does not indicate whether or not S1 has friction. It's clear that S2 has friction. I would suggest solving it both ways.

.
 
SammyS said:
The normal force for S2 is obviously mg.

What is the normal force for S1, when the block is on the ramp?

I do have a question about the problem. It does not indicate whether or not S1 has friction. It's clear that S2 has friction. I would suggest solving it both ways.

.

The professor told me that it is the same coefficient for both
 
SammyS said:
OK !

What is the normal force, when the block is on the ramp?

mg cos 30.3?
 
alfredo24pr said:
mg cos 30.3?

Yes, that's right. (Make sure that's 30.3° , not 30° & 30 minutes. The latter is sometimes written 30:30° or 30°30'. 30° & 30 minutes = 30.5° )

The frictional force is μ times the normal force & is in the direction opposite the direction of the motion.