Question

A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has force constant 815 N/m . The coefficient of kinetic friction between the floor and the block is μk=0.40. The block and spring are released from rest and the block slides along the floor.
Required:
What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0100 m.)

Answer 1

Answer:

v = 0.41 m/s

Explanation:

• In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
• At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
• So, we can write the following general equation, taking the initial and final values of the energies:

       $\Delta K + \Delta U = W_{ffr} (1)$

• Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
• ⇒ Kf = 1/2*m*vf²  (2)
• The change in the potential energy, can be written as follows:

       $\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

• Regarding the work done by the force of friction, it can be written as follows:

       $W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

• Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
• Fn = Fg= m*g (5)
• Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        $\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$

        $\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)$

• Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    $\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)$

• $v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)$