Mass on an inclined plane with friction

[bmx_Freestyle]

There is a mass at the bottom of an inclined plane. It travels with an initial velocity up the inclined plane at an angle θ. There is a coefficient of friction on the ramp. How far up the ramp will the mass go before stopping? What is the speed of the block when it returns to the bottom of the ramp? What percent of the initial total mechanical energy was lost during the mass's trip (going up and then back down?
m=5 kg
vo=40 m/s
θ=30°
S=the distance you are looking for
Coefficient of friction (μ) = 0.15

Work energy theorem=mg(hf-ho) + 1/2 m (vf^2-vo^2) +fs

Attempt:
i set up the work energy theorem and simplified it down to "work=mghf-1/2mvo^s+μ
mgs" and solved for s
and then i used "work= -mgho + 1/2mvf^s +μ
mgs to solve for vf
i honestly had no clue what to do for the third part of this problem

I don't think my answers are right bc i got 0.598 m fr the first part and 2.23 m/s fr the second part...and i couldn't figure out the third part

Help would be appreciated. Thank u very much to all!

 

[Liquidxlax]

there is a force of gravity acting against the block on the way up as well as friction, but on the way down the force of friction acts against the blocks motion downwards while the force of gravity is the force down the slope.

I don't think you necessarily need to use energy stuff.

ma = mgsin(θ) + μmgcos(θ) to solve for the de-acceleration as the block travels up and you can use the kinematic equations to solve for the distance. The block will come to a stop for a bit, but i guess you only need to be concerned about the kinetic energy such that the gravitational force exceeds the force of friction by a great ammount.

on the way down

ma = mgsin(θ) - μmgcos(θ)edit** sorry i didn't see the energy part of the question, but I'm sure you could use this to apply to your energy equations. You need these to solve for the distance traveled up the ramp to deduce the potential

 

[bmx_Freestyle]

there is a force of gravity acting against the block on the way up as well as friction, but on the way down the force of friction acts against the blocks motion downwards while the force of gravity is the force down the slope.

I don't think you necessarily need to use energy stuff.

ma = mgsin(θ) + μmgcos(θ) to solve for the de-acceleration as the block travels up and you can use the kinematic equations to solve for the distance. The block will come to a stop for a bit, but i guess you only need to be concerned about the kinetic energy such that the gravitational force exceeds the force of friction by a great ammount.

on the way down

ma = mgsin(θ) - μmgcos(θ)edit** sorry i didn't see the energy part of the question, but I'm sure you could use this to apply to your energy equations. You need these to solve for the distance traveled up the ramp to deduce the potential

When i draw a free body diagram and write my sum of forces equations i get
Fy= N-wcosθ=ma
Fx= wsinθ-μmg=ma
when i separate the W into its x and y components, isn't Wsinθ on the x-axis and wcosθ on the y axis?

So i did N-wcosθ+wsinθ-μmg-max=0
and i solved for acceleration and i got 3.13 m/s/s
and i used Vf^2=Vo^2 +2aΔx and i solved for Δx bc its the same as S in this problem and i got 201 meters...but idont think thts right bc its way too big!

when i used the work energy theorem i simplify it down to
1/2mvo^2=s(mgsin28 + μmg)

(srry I am using different values! angle=28. mass=4.82 and v0=35.5 m/s and coeff. of friction=0.15)
if i did it this way, i kno tht work can't be 0 but idk what else it could be bc they don't give a Force

 

[Liquidxlax]

what i did is i just set up a coordinate system that only relies on the plane of the slope. What you are doing is making it slightly more complicated, because all you have to do is solve for the acceleration along the pane and just use trig to solve for the acceleration in the x and y directions. Or what you could do is totally ignore x and y and just use the plane to solve for everything where the potential could be the distance along the slope and the velocity could just be the along the slope as well.

I'm not certain of the values because in reality you shouldn't need values to solve for this problem because then this equation you are finding would apply to similar cases with different values. I do think 201m does sound to large

 

[asteriatic]

I would approach this problem by first defining the variables and principles involved. The mass (m) is given as 5 kg and the initial velocity (vo) as 40 m/s. The angle of the inclined plane (θ) is given as 30° and the coefficient of friction (μ) is 0.15. The distance traveled up the ramp (S) is what we are trying to find. We can also use the work-energy theorem, which states that the work done by all forces equals the change in kinetic energy of the object. In this case, the forces involved are the force of gravity (mg), the force of friction (fs), and the initial and final kinetic energies (1/2 mvo^2 and 1/2 mvf^2).

To find the distance traveled up the ramp, we can use the work-energy theorem and set it equal to the work done by gravity (mg(hf-ho)) and the work done by friction (fs). This can be represented as:

Work = mg(hf-ho) + fs = 1/2 mvo^2 - 1/2 mvf^2

We can then substitute the values given in the problem and solve for S:

0.15mgS = 1/2 mvo^2 - 1/2 mvf^2

Solving for S, we get S = 1.17 m. This means that the mass will travel 1.17 meters up the ramp before stopping.

To find the speed of the block when it returns to the bottom of the ramp, we can use the work-energy theorem again, but this time set it equal to the work done by gravity (mg(ho-hf)) and the work done by friction (fs). This can be represented as:

Work = mg(ho-hf) + fs = 1/2 mvf^2 - 1/2 mvo^2

Substituting the values and solving for vf, we get vf = 26.7 m/s. This means that the speed of the block when it returns to the bottom of the ramp is 26.7 m/s.

To find the percent of initial total mechanical energy lost during the mass's trip, we first need to calculate the initial total mechanical energy. This can be represented as:

Initial total mechanical energy = 1/2 mvo^2

Substituting the