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Physics Homework Help???

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A blcok of mass 6.00 kg from rest down a 30 degree frictionless incline. At the bottom of the incline it is stopped by a spring with k=3.00 x 10^(4) N/m/ The blcok slides 2.00 m from the point of release to the point where it comes to rest against the spring. When the blcok comes to rest, how far has the spring been compressed?
asked Oct 28, 2014 in PHYSICS by heather Apprentice

1 Answer

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Mass os block m = 6 kg

Frictionless inclined angle θ = 30o

Spring constant k = 3.00 x 104 N/m

Distance of block slides from the point of release to the point where it comes to rest against the spring = 2.00 m

compressed distance of spring d = ?

In describing the initial potential energy, what do we want to choose as the reference position.

Choose to make the potential energy zero when the mass stops after compressing the spring.

That is, We need to choose the reference point for the potential energy is the location at which the potential energy is zero is

 2-d m

From diagram

h =  2  sin 30o

h =  2  (0.5)

h = 1

The Mechanical Energy (E) of a system is the sum of its Kinetic (K) and Potential (U) energies.

Total initial mechanical energy Ei = Ki + Usi + Ugi

Ki = Final Kinetic Energy = ½mv²

Ugi = Final Gravitational Potential Energy = mgh

Usi = Final Spring Potential Energy = ½kx²

Ei = Ki + Usi + Ugi

Ei = 0 + 0 + mgh

Ei = 6*9.8*1

Ei = 58.8

Total final mechanical energy Ef = Kf + Usf + Ugf

Kf = Final Kinetic Energy

Ugf = Final Gravitational Potential Energy = mgh

Usf = Final Spring Potential Energy = ½kx²

Ef = Kf + Usf + Ugf

Ef = 0 + ½kd² + 0

Ef = (3000)d²

Ef = 3000d²

From Conservation of Energy : Ei = Ef

58.8 = 3000d²

58.8/3000 = d²

d² = 0.0196 m

d = 0.00384 m

The compressed distance of spring d = 0.00384 m

 

answered Oct 28, 2014 by lilly Expert

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