Physics Spring Problem

Question

Another spring, another hill, this time a bit more geometric (see figure). A block (m = 0.41 kg) is pushed 0.65 m against a spring and released from rest. The block then travels over an icy patch of ground (assumed frictionless) before reaching a hill inclined θ = 33.6^o above the horizontal. The slope of the hill is not frictionless, but rather has a force of friction between the block and the hill of 2.01 N. The block comes to rest when it is h = 4.2 m above its starting point. What is the spring constant of the spring?

Answer 1

Since the figure was not included let the spring be at point A the base of the ramp be point B and the point where the mass stops be point C. Hence U_sA = k(.65)²/2 U_gC = .41* 9.8 * 4.2 K_A =K_c =0 and W_fBC = (2.01*4.2)/sin 33.6 so U_sA = .2113 k U_gC = 16.8756 and W_fBC = 15.255.

Using conservation of energy K_C - K_A -U_gC +U_sA = -W_fBC hence 0 - 0 - 16.8756 + .2113 k = -15.255 solving for k .2113 k = 1.6205 k= 7.67 N-m

Answer 2

Givens:

m=0.41kg

Δx=0.65m

θ=33.6º

F_f=2.01N

h=4.2m

Formulas:

PE_s=1/2kΔx²

PE_g=mgh

W=F•d

Draw your triangle of your incline, how does the height of the block compare with the distance the block traveled along the incline? This will be your "d" in your Work formula

sinθ=h/d

d=h/sinθ

d=7.59m

Now think about your energy relationships. What energy do you have in the beginning? Potential Energy stored in the spring

What energy do you have at the end? Gravitational Potential Energy

What energy was removed from the system along the way (or used to do work)? The work done opposing friction

Therefore

PE_s-W_f=PE_g

1/2kΔx²-F_fd=mgh

1/2kΔx²=mgh+F_fd

kΔx²=2(mgh+F_fd)

k=2(mgh+F_fd)/Δx²

k=2((0.41)(9.81)(4.2)+(2.01)(7.59))/(0.65)²

k=152.2N/m