Question

A spring (70 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41 ∘ with respect to the horizontal. The spring is then released.
Required:
a. If the mass is not attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?
b. If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?
c. Now the incline has a coefficient of kinetic friction μk. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction μk?

Answer 1

Answer:

A) l = 0.619m

B) l = 0.596m

C) μ = 0.314

Explanation:

The data given is:

k = 70 N/m

x = 0.5 m

m = 2.2 kg

θ = 41°

(FIGURES FOR EACH PART ARE ATTACHED AT THE BOTTOM. CONSULT THEM FOR BETTER UNDERSTANDING)

Part A

Gain in Gravitational Potential Energy = Loss in Elastic Potential Energy

mgh = (1/2)kx²

(2.2)(9.8)h = (1/2)(70)(0.5)²

h = 0.406 m

sinθ = h/l

l = h / sinθ

l = 0.406/sin41

l = 0.619m

Part B

Loss in Elastic Potential Energy in compressed spring = Gain in Gravitational Potential Energy + Gain in Elastic Potential Energy in stretched spring

(1/2)kx² = mgh + (1/2)k(l - 0.5)²

(1/2)(70)(0.5)² = (2.2)(9.8)(l·sin41)) + (1/2)(70)(l² - l + 1/4)

8.75 = 14.15(l) + 35(l²) - 35(l) + 8.75

35(l²) -20.85(l) = 0

l = 0.596m

Part C

Loss in Elastic Potential Energy = Gain in Gravitational Potential Energy + Work done against friction

(1/2)kx² = mgh + Fd

(1/2)kx² = mg(dsinθ) + μRd

(1/2)kx² = mg(dsinθ) + μ(mg · cosθ)d

(1/2)kx² = mgd (sinθ + μ(cosθ))

(1/2)(70)(0.5)² = (2)(9.8)(0.5) (sin41 + μcos41)

8.75 = 6.43 + 7.4μ

μ = 0.314

Answer 2

Answer:

a) The mass moves a distance of 0.625 m up the slope before coming to rest

b) The distance moved by the mass when it is connected to the spring is 0.6 m

c) $\mu = 0.206$

Explanation:

Spring constant, k = 70 N/m

Compression, x = 0.50 m

Mass placed at the free end, m = 2.2 kg

angle, θ = 41°

Potential Energy stored in the spring, $PE= 0.5 kx^2$

$PE = 0.5 * 70 * 0.5^2\\PE = 8.75 J$

According to the principle of energy conservation

PE = mgh

8.75 = 2.2 * 9.8 * h

h = 0.41

If the mass moves a distance d from the spring

sin 41 = h/d

sin 41 = 0.41/d

d = 0.41/(sin 41)

d = 0.625 m

The mass moves a distance of 0.625 m up the slope before coming to rest

b) If the mass is attached to the spring

According to energy conservation principle:

Initial PE of spring = Final PE of spring + PE of block

$0.5kx_1^2 = 0.5kx_2^2 + mgh\\x_2 = d - x_1 = d - 0.5\\h = d sin 41\\0.5*70*0.5^2 = 0.5*70*(d-0.5)^2 + 2.2*9.8*d*sin41\\8.75 = 35(d^2 - d + 0.25) + 14.15d\\8.75 = 35d^2 - 35d + 8.75 + 14.15d\\35d^2 = 20.85d\\d = 0.6 m$

The distance moved by the mass when it is connected to the spring is 0.6 m

3) The spring potential is converted to increased PE and work within the system.

mgh = Fd + 0.5kx²...........(1)

d = x , h = dsinθ

kinetic friction force , F = μmgcosθ

mgdsinθ + μmg(cosθ)d = 0.5kd²

mgsinθ + μmgcosθ = 0.5kd

sinθ + μcosθ = kd/(2mg)

$\mu = \frac{\frac{kd}{2mg} - sin\theta}{cos\theta} \\\\\mu = \frac{\frac{70*0.5}{2*2.2*9.8} - sin41}{cos41} \\\\\mu = 0.206$