Respuesta :
Answer:
a)
[tex]0.26[/tex] m
b)
[tex] 8.2[/tex] J
Explanation:
a)
Consider the motion of the block between equilibrium point and the point where it comes to a stop
[tex]KE_{o}[/tex] = initial Kinetic energy of the block at equilibrium position = 20.7 J
[tex]k[/tex] = spring constant of the spring = 432 Nm⁻¹
[tex]f[/tex] = frictional force acting on the block = 24 N
[tex]x[/tex] = stretch in the spring caused before it stops
Using conservation of energy at initial equilibrium point and stopping point
Kinetic energy at equilibrium position = Spring energy at the stopping point + magnitude of work done by frictional force
[tex]KE_{o} = (0.5)kx^{2} + fx[/tex]
inserting the values
[tex]20.7 = (0.5)(432)x^{2} + (24)x[/tex]
[tex](216)x^{2} + (24)x - 20.7 = 0[/tex]
[tex]x = 0.26[/tex] m
b)
Consider the motion of the block between equilibrium point and the same point when it returns
[tex]KE_{o}[/tex] = initial Kinetic energy of the block at equilibrium position = 20.7 J
[tex]KE_{o}[/tex] = final Kinetic energy of the block at equilibrium position while returning
[tex]f[/tex] = frictional force acting on the block = 24 N
[tex]d[/tex] = distance traveled by the block = [tex]x + x[/tex] = [tex]2x[/tex]
Using conservation of energy at initial equilibrium point and stopping point
Kinetic energy at equilibrium position = Kinetic energy at equilibrium while returning + magnitude of work done by frictional force
[tex]KE_{o} = KE_{o} + 2 fx[/tex]
inserting the values
[tex]20.7 = KE_{o} + 2 (24)(0.26)[/tex]
[tex]20.7 = KE_{o} + 12.5[/tex]
[tex]KE_{o} = 8.2[/tex] J
