Question

A massless spring with force constant 602 N/m is fastened at its left end to a vertical wall, as shown below. Initially, the 6 kg block and 3 kg block rest on a horizontal surface with the 6 kg block in contact with the spring (but not compressing it) and with the 3 kg block in contact with the 6 kg block. The 6 kg block is then moved to the left, compressing the spring a distance of 0.6 m, and held in place while the 3 kg block remains at rest as shown below. The acceleration of gravity is 9.8 m/s2
A) Determine the elastic energy U stored in the compressed spring. Answer in units of J ?

B) The 6 kg block is then released and accelerates to the right, toward the 3 kg block. The surface is rough and the coefficient of friction between each block and the surface is 0.3 . The two blocks collide, stick together, and move to the right. Remember that the spring is not attached to the 6 kg block. Find the speed of the 6 kg block just before it collides with the 3 kg block. ? Answer in units of m/s.

C) Find the final speed of both blocks (stuck together) just after they collide. Answer in units of m/s?

D) Find the horizontal distance the blocks move before coming to rest. Answer in units of m ?

Answer 1

Answer:

A) Ke = 108 J , B)  v = 10.24 m/s , C)  v₂ = 6.83 m/s and D)    $x_{f}$ = 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)

    $Em_{f}$= K = ½ m v²

How is friction force  

    $W_{fr}$ = ΔDEm = $Em_{f}$ - 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

    $p_{f}$ = (m₁ + m₂) v²

   p₀ = $p_{f}$

   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

    $W_{fr}$ = ΔDEm = Em₃ -Em₂

    -fr $x_{f}$ = 0 - ½ (m₁ + m₂) v₂²

We use the friction force equation of the above

    -μ (m₁ + m₂) g xf = - ½ (m₁ + m₂) v₂²

     $x_{f}$ = ½ v₂² / μ g

    $x_{f}$ = ½ 6.83² / (0.3 9.8)

   $x_{f}$ = 7.93 m