How to Calculate Coefficient of Kinetic Friction?

[HardestPart]

Homework Statement

When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.07 m/s. The mass stops a distance S2 = 2.5 m along the level part of the slide. The distance S1 = 1.13 m and the angle q = 30.50°. Calculate the coefficient of kinetic friction for the mass on the surface.

Homework Equations

http://capa2.cc.huji.ac.il/res/msu/physicslib/msuphysicslib/13_EnergyConservation/graphics/prob27a_MechEnWFriction.gif

The Attempt at a Solution

I tried solving this question by:
H=1.13sin30.5
mgH+1/2mv^2-(s1+s2)u*mgcosa=0
9.8*0.5735+1/2*(2.07^2)-3.63u*9.8cos30.5=0
5.6203+2.14245=30.651u
u=0.2532
I get wrong answer all the time
can someone tell me where i went wrong?

I have another question which I have no clue hot to solve it:

The left side of the figure shows a light (`massless') spring of length 0.290 m in its relaxed position. It is compressed to 71.0 percent of its relaxed length, and a mass M= 0.250,kg is placed on top and released from rest (shown on the right).
http://capa2.cc.huji.ac.il/res/msu/physicslib/msuphysicslib/13_EnergyConservation/graphics/prob24_CompSpring.gif
The mass then travels vertically and it takes 1.50 s for the mass to reach the top of its trajectory. Calculate the spring constant, in N/m. (Use g=9.81 m/s2). Assume that the time required for the spring to reach its full extension is negligible.

Can someone tell me how to do this question?

 

[PhanthomJay]

Homework Statement

When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.07 m/s. The mass stops a distance S2 = 2.5 m along the level part of the slide. The distance S1 = 1.13 m and the angle q = 30.50°. Calculate the coefficient of kinetic friction for the mass on the surface.

Homework Equations

http://capa2.cc.huji.ac.il/res/msu/physicslib/msuphysicslib/13_EnergyConservation/graphics/prob27a_MechEnWFriction.gif

The Attempt at a Solution

I tried solving this question by:
H=1.13sin30.5
mgH+1/2mv^2-(s1+s2)u*mgcosa=0
9.8*0.5735+1/2*(2.07^2)-3.63u*9.8cos30.5=0
5.6203+2.14245=30.651u
u=0.2532
I get wrong answer all the time
can someone tell me where i went wrong?

why sure, in that last term you have s1(u)mgcos theta, which is fine, but then the friction force over the 's2' length is over a level surface, so there's no cos part, its just s2(u)(mg)

I have another question which I have no clue hot to solve it:

The left side of the figure shows a light (`massless') spring of length 0.290 m in its relaxed position. It is compressed to 71.0 percent of its relaxed length, and a mass M= 0.250,kg is placed on top and released from rest (shown on the right).
http://capa2.cc.huji.ac.il/res/msu/physicslib/msuphysicslib/13_EnergyConservation/graphics/prob24_CompSpring.gif
The mass then travels vertically and it takes 1.50 s for the mass to reach the top of its trajectory. Calculate the spring constant, in N/m. (Use g=9.81 m/s2). Assume that the time required for the spring to reach its full extension is negligible.

Can someone tell me how to do this question?

Someone can help if you show an attempt at a solution, but beyond that, please post separate questions in separate posts, to avoid the chaos that develops when multiple questions on different topics are posted together.

 

[HardestPart]

I tried to solve the question ne more time as you told me,but am still getting a wrong answer

mgH+1/2mv^2-s1umgcosa-s2umg=0

is the equation is correct?

 

[PhanthomJay]

I tried to solve the question ne more time as you told me,but am still getting a wrong answer

mgH+1/2mv^2-s1umgcosa-s2umg=0

is the equation is correct?

Looks good, if I'm interpreting the problem correctly. So I get 5.62+2.14 -u(9.54 + 24.5)= 0
Solve u = 0.23, did you have a math error?