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dgl7
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1.
A spring with a spring-constant 3.6 N/cm is compressed 39 cm and released. The 9 kg mass skids down the frictional incline of height 36 cm and inclined at a 20◦ angle. The acceleration of gravity is 9.8 m/s^2. The path is frictionless except for a distance of 0.6 m along the incline which has a coefficient of friction of 0.4 .
What is the final velocity of the mass?
2. RELEVANT EQUATIONS
Ko+Uo+Wnc=U+K
kinematic equations
ma=kx
3. MY ATTEMPT
I tried kx=ma-->
(3.6)(.39)=(9)a
a=.156
I put that into the kinematic equation v^2 = vo^2 + 2a(X - Xo)
v^2=(0)^2+2(.156)(.39)
got v=.3488266045 m/s (note: this would be vo for the overall equation and v only for the compression of the spring)
I put that into Ko in the Ko+Uo+Wnc=U+K equation-->
.5(9)(.3488266045)^2+(9)(9.8)(.36)+[(.4)(9*9.8*cos20)](.6)(cos20)=0+.5(9)v^2
solved for v and got 3.366216946 m/s but that isn't right...I think there is something wrong with the nonconservative work, but I'm not sure what...
A spring with a spring-constant 3.6 N/cm is compressed 39 cm and released. The 9 kg mass skids down the frictional incline of height 36 cm and inclined at a 20◦ angle. The acceleration of gravity is 9.8 m/s^2. The path is frictionless except for a distance of 0.6 m along the incline which has a coefficient of friction of 0.4 .
What is the final velocity of the mass?
2. RELEVANT EQUATIONS
Ko+Uo+Wnc=U+K
kinematic equations
ma=kx
3. MY ATTEMPT
I tried kx=ma-->
(3.6)(.39)=(9)a
a=.156
I put that into the kinematic equation v^2 = vo^2 + 2a(X - Xo)
v^2=(0)^2+2(.156)(.39)
got v=.3488266045 m/s (note: this would be vo for the overall equation and v only for the compression of the spring)
I put that into Ko in the Ko+Uo+Wnc=U+K equation-->
.5(9)(.3488266045)^2+(9)(9.8)(.36)+[(.4)(9*9.8*cos20)](.6)(cos20)=0+.5(9)v^2
solved for v and got 3.366216946 m/s but that isn't right...I think there is something wrong with the nonconservative work, but I'm not sure what...
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