When a 0.111 kg mass is suspended at rest from a certain spring, the spring stretches 3.90 cm. Find the instantaneous acceleration of the mass when it is raised 6.20 cm, compressing the spring 2.30 cm.
from the static positon
k=mg/x
now, knowing k
a=g+kx/mass
To find the instantaneous acceleration of the mass, we can use Hooke's Law and the concept of energy conservation.
Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from equilibrium. Mathematically, this can be written as:
F = -kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from equilibrium.
In this problem, we are given the mass of the object (m = 0.111 kg), the displacement of the spring when the object is at rest (x1 = 3.90 cm), and the compressional displacement of the spring (x2 = -2.30 cm). We need to find the instantaneous acceleration.
First, let's convert the given displacements to meters:
x1 = 3.90 cm = 3.90/100 = 0.039 m (stretching the spring)
x2 = -2.30 cm = -2.30/100 = -0.023 m (compressing the spring)
Now, we need to find the spring constant, k. We can do this by using Hooke's Law, setting the force equal to the weight of the object:
mg = kx1
Where g is the acceleration due to gravity (9.8 m/s^2). Rearranging the equation, we get:
k = mg / x1
Substituting the given values, we have:
k = (0.111 kg)(9.8 m/s^2) / 0.039 m ≈ 28.38 N/m
Now, we can find the potential energy of the system when the spring is compressed (U2) and compare it with the potential energy of the system when the spring is at rest (U1):
U1 = (1/2)kx1^2
U2 = (1/2)kx2^2
The difference between U2 and U1 represents the potential energy change of the system, which can be equated to the kinetic energy of the mass:
U2 - U1 = (1/2)mv^2
Where v is the velocity of the mass.
Since we are interested in finding the acceleration, we can differentiate both sides of the equation with respect to time:
d/dt (U2 - U1) = d/dt ((1/2)mv^2)
The left side of the equation represents the change in potential energy with respect to time, which is equal to the force exerted by the spring multiplied by the displacement rate of the mass:
F = -kx2 (since the spring is compressed, the displacement is in the opposite direction)
dU/dt = -kx2(dx2/dt)
The right side of the equation represents the derivative of kinetic energy with respect to time, which is equivalent to the mass multiplied by the acceleration of the mass:
(mv)dv/dt = m(dv/dt)
Setting these two sides equal to each other, we have:
-kx2(dx2/dt) = m(dv/dt)
Rearranging and substituting the given values, we can solve for the instantaneous acceleration (dv/dt):
(dv/dt) = -kx2 / m
Substituting the values:
(dv/dt) = (-28.38 N/m)(-0.023 m) / 0.111 kg
Evaluating this expression will give us the instantaneous acceleration of the mass when it is raised 6.20 cm, compressing the spring 2.30 cm.