Three-Part Physics Question - help and answers appreciated

Question

Answer 1

Let

m = 0.5 kg be the mass of the block,

k = 1200 N/m be the spring constant,

x = 15 cm = 0.15 m be the compression of the spring,

θ = 30° be the incline of the ramp.

(a)

The energy of the spring

kx²/2

gets converted into the potential energy of the block at distance d up the ramp.

The height at the distance d is

h = sin(θ)d.

Thus the statement of conservation of energy is

kx²/2 = sin(θ)dmg

From that

d = kx²/(2sin(θ)mg)

Substituting actual numbers

d = 1200 × 0.15² / (2 × sin(30°) × 0.5 × 9.81) = 5.5 m

(b)

The energy of the spring

kx²/2

gets converted into the potential energy of the block at distance d/2 up the ramp

plus the work, W, of friction.

Thus the statement of conservation of energy is

kx²/2 = sin(θ)dmg/2 + W

W = kx²/2 - sin(θ)dmg/2 = kx²/4

W = 1200 × 0.15² / 4 = 6.75 J

(c)

Let

F be the force of friction,

f be the coefficient of friction on the ramp.

F is related to the work by

W = Fd/2

F = 2W/d

And F is related to the block and the coefficient of friction by

F = cos(θ)mgf

Thus

f = F/(cos(θ)mg)

= 2W/(cos(θ)dmg)

= (kx²/2) / (cos(θ)kx²mg/(2sin(θ)mg)) = tan(θ) = tan(30°) = √3/3