Answer:
A) Ke = 108 J
, B) v = 10.24 m/s
, C) v₂ = 6.83 m/s and D) = 7.93 m
Explanation:
rA) let's use the elastic energy equation
Ke = ½ k x²
Let's calculate
Ke = ½ 602 0.6²
Ke = 108 J
r.b) in this case the surface has friction, so it performs a work that opposes the movement and must be equal to the variation of the mechanical energy. Let's write the mechanical energy in two points
initial maximum compression
Em₀ = Ke = ½ k x²
final. just when you touch the small block; at this point the elongation of the zero spring (x = 0)
= K = ½ m v²
How is friction force
= ΔDEm = - Em₀
-fr x = ½ m v² - ½ k x²
Let's use Newton's second law to find the force of friction
fr = μ N
Equation on the y axis
N- W = 0
N = mg
fr = μ m g
Let's replace in the equation of work and energy
- (μ mg) x = ½ m v² - ½ k x²
Calculation of the speed of the large block
v² = (½ k x² - μ mg x) 2/m
v = √ (k / m x² - 2 μ g x)
v = √ (602/6 0.6² - 2 0.3 9.8 0.6) = √ (108.36 -3.53)
v = 10.24 m/s
rC) For this part we use the conservation of the moment, here we must define a system formed by the two bodies, for this system the forces during the internal crash, so that the moment is conserved
m₁ = 6 kg
m₂ = 3 kg
before the crash
p₀ = m₁ v + 0
After the crash, before starting to move
= (m₁ + m₂) v²
p₀ =
m₁ v = (m₁ + m₂) v₂
v₂ = v m₁ / (m₁ + m₂)
v₂ = 10.24 6 / (6 +3)
v₂ = 6.83 m / s
This is the speed of the two blocks together
rD) Here again we use energy conservation
starting point
Just when the blocks already collided and have the speed v2, but they have not begun to move
Em₂ = K = ½ (m₁ + m₂) v₂²
final point when the blocks stop
Em₃ = 0
Let's write the work and energy equation
= ΔDEm = Em₃ -Em₂
-fr = 0 - ½ (m₁ + m₂) v₂²
We use the friction force equation of the above
-μ (m₁ + m₂) g xf = - ½ (m₁ + m₂) v₂²
= ½ v₂² / μ g
= ½ 6.83² / (0.3 9.8)
= 7.93 m