Physics Help????

Question

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?

Answer 1

Mass os block m = 6 kg

Frictionless inclined angle θ = 30^o

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 30^o

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 E_i = K_i + U_si + U_gi

K_i = Final Kinetic Energy = ½mv²

U_gi = Final Gravitational Potential Energy = mgh

U_si = Final Spring Potential Energy = ½kx²

E_i = K_i + U_si + U_gi

E_i = 0 + 0 + mgh

E_i = 6*9.8*1

E_i = 58.8

Total final mechanical energy E_f = K_f + U_sf + U_gf

K_f = Final Kinetic Energy

U_gf = Final Gravitational Potential Energy = mgh

U_sf = Final Spring Potential Energy = ½kx²

E_f = K_f + U_sf + U_gf

E_f = 0 + ½kd² + 0

E_f = (3000)d²

E_f = 3000d²

From Conservation of Energy : E_i = E_f

58.8 = 3000d²

58.8/3000 = d²

d² = 0.0196 m

d = 0.00384 m

The compressed distance of spring d = 0.00384 m