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
\