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) [tex]\mu = 0.206[/tex]
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, [tex]PE= 0.5 kx^2[/tex]
[tex]PE = 0.5 * 70 * 0.5^2\\PE = 8.75 J[/tex]
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
[tex]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[/tex]
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)
[tex]\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[/tex]
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)
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
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
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
