A 10.7 g mass is attached to a horizontal spring with a spring constant of 12.6 N/m and released from rest with an amplitude of 37.8 cm.
What is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless?
To find the speed of the mass when it is halfway to the equilibrium position, we can use the principle of conservation of mechanical energy.
The total mechanical energy of the system is conserved, and it can be expressed as the sum of the kinetic energy and the potential energy:
E = KE + PE
Initially, when the mass is at the maximum displacement, all the energy is potential energy:
E_initial = PE_initial
At halfway to the equilibrium position, the mass has converted half of its potential energy into kinetic energy:
E_midway = (1/2)KE + (1/2)PE
Finally, when the mass is at the equilibrium position, all the energy is kinetic energy:
E_final = KE_final
Since the surface is frictionless, there is no energy loss due to friction, and the initial mechanical energy is equal to the final mechanical energy:
E_initial = E_final
Therefore:
PE_initial = KE_final
Let's calculate the initial potential energy and set it equal to the final kinetic energy.
The potential energy of a spring is given by the formula:
PE = (1/2)kx^2
where k is the spring constant and x is the displacement from the equilibrium position.
Since the amplitude is given as 37.8 cm, the maximum displacement x is half of the amplitude:
x = 37.8 cm / 2 = 18.9 cm = 0.189 m
The potential energy at the maximum displacement is:
PE_initial = (1/2)kx^2 = (1/2)(12.6 N/m)(0.189 m)^2
Now, we need to calculate the final kinetic energy when the mass is halfway to the equilibrium position.
The potential energy at halfway to the equilibrium position is:
PE_midway = (1/2)k(0.189 m / 2)^2 = (1/8)kx^2
Since the mass has converted half of its potential energy into kinetic energy:
KE_midway = (1/2)PE_midway = (1/16)kx^2
We can set these two equations equal to each other:
PE_initial = KE_midway
(1/2)(12.6 N/m)(0.189 m)^2 = (1/16)(12.6 N/m)(0.189 m)^2
Simplifying:
(1/2) = (1/16)
Since both sides are equal, we cancel out common factors:
1 = 1
This means that the initial potential energy is equal to the final kinetic energy.
Therefore, the speed of the mass when it is halfway to the equilibrium position is the same as the speed at the maximum displacement.
To find the speed, we can equate kinetic energy to (1/2)mv^2, where m is the mass and v is the speed:
(1/2)mv^2 = (1/2)kx^2
We can rearrange the equation to solve for v:
v = sqrt((kx^2) / m)
Substituting the given values:
v = sqrt((12.6 N/m)(0.189 m)^2 / 0.0107 kg)
Calculating:
v ≈ sqrt(0.146 N m / 0.0107 kg)
v ≈ sqrt(13.63 m^2/s^2)
v ≈ 3.69 m/s
Therefore, the speed of the mass when it is halfway to the equilibrium position is approximately 3.69 m/s.
To find the speed of the mass when it is halfway to the equilibrium position, we need to calculate the potential energy and then convert it into kinetic energy.
To begin, let's determine the potential energy of the mass when it is halfway to the equilibrium position. At this point, the mass has moved an amplitude of 37.8 cm / 2 = 18.9 cm or 0.189 m.
The potential energy of a mass-spring system can be calculated using the formula:
Potential Energy = (1/2) * k * x^2
Where:
k is the spring constant (12.6 N/m)
x is the displacement from the equilibrium position (0.189 m)
Potential Energy = (1/2) * (12.6 N/m) * (0.189 m)^2
= 0.22247 J
Now, let's convert this potential energy into kinetic energy, assuming no energy loss due to friction or other factors. We know that the total mechanical energy is conserved, so the potential energy at the halfway point is equal to the kinetic energy.
Kinetic Energy = Potential Energy
(1/2) * m * v^2 = 0.22247 J
Rearranging the equation, we can solve for the velocity (v):
v^2 = (2 * 0.22247 J) / m
Substituting the mass of the object (10.7 g) into kg (1 g = 0.001 kg):
v^2 = (2 * 0.22247 J) / (10.7 g * 0.001 kg/g)
v^2 = 41.6542 m^2/s^2
Taking the square root of both sides, we can find the velocity (v):
v = √(41.6542 m^2/s^2)
v ≈ 6.46 m/s
Therefore, the speed of the mass when it is halfway to the equilibrium position is approximately 6.46 m/s.
omega = sqrt (k/m)
= sqrt (12.6/.0107) = 34.3 radians/s
x = .378 cos (omega t)
v = -.378(34.3) sin (omega t)
when x = .378/2 cos(omega t) = .5
then omega t = 60 degrees = pi/3 radians
then sin (omega t) = .866
(I am ignoring signs here, we need speed not velocity)
v = .378 * 34.3 * .866
= 11.2 m/s