A spring of negligible mass stretches 2.20 cm from its relaxed length when a force of 10.50 N is applied. A 1.100 kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x=5.0 cm and released from rest at t=0.
(d) What is the maximum velocity of the system?
To find the maximum velocity of the system, we first need to determine the equilibrium position of the spring. Since the spring stretches 2.20 cm when a force of 10.50 N is applied, we can use Hooke's Law to find the spring constant.
Hooke's Law states that the force applied to a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be written as:
F = -kx
Where F is the force applied, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, we have the force applied (10.50 N) and the displacement (2.20 cm or 0.0220 m). So we can rearrange the equation to solve for the spring constant:
k = -F/x = -10.50 N / 0.0220 m = -477.3 N/m
Now that we know the spring constant, we can find the potential energy of the system at the maximum displacement (x = 5.0 cm or 0.050 m). The potential energy stored in the spring is given by:
E = (1/2)kx^2
Substituting the values, we get:
E = (1/2)(-477.3 N/m)(0.050 m)^2 = -0.5966 J
Since total mechanical energy is conserved, the potential energy at the maximum displacement is converted entirely into kinetic energy at the maximum velocity. Therefore, we can equate the potential energy to the kinetic energy:
E = (1/2)mv^2
Solving for v, we get:
v = sqrt(2E/m) = sqrt(2(-0.5966 J) / 1.100 kg)
Calculating this expression, we find:
v ≈ 0.670 m/s
So the maximum velocity of the system is approximately 0.670 m/s.