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:
[tex]\Delta K + \Delta U = W_{ffr} (1)[/tex]
- 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:
[tex]\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)[/tex]
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:
[tex]W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)[/tex]
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:
[tex]\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)[/tex]
[tex]\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)[/tex]
- Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:
[tex]\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)[/tex]
- [tex]v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)[/tex]
