A 200 g object on a frictionless, horizontal lab table is pushed against a spring of force constant 35.0 N/cm and then released. Just before the object is released, the spring is compressed 11.0 cm.
How fast is the object moving when it has gained half of the spring's original stored energy?
To determine the speed of the object when it has gained half of the spring's original stored energy, we can use the principle of conservation of mechanical energy.
The potential energy stored in a spring is given by the equation:
PE = (1/2)kx^2
Where:
PE is the potential energy stored in the spring,
k is the force constant of the spring,
x is the displacement of the spring from its equilibrium position.
In this case, just before the object is released, the spring is compressed by 11.0 cm. So, the potential energy stored in the spring is:
PE = (1/2)(35.0 N/cm)(0.11 m)^2
PE = 0.04255 J
Since the object gains half of this energy, the total mechanical energy gained by the object is:
Total Mechanical Energy = (1/2)PE
Total Mechanical Energy = (1/2)(0.04255 J)
Total Mechanical Energy = 0.02127 J
To determine the speed of the object, we can equate the mechanical energy gained by the object to its kinetic energy:
Total Mechanical Energy = (1/2)mv^2
Where:
m is the mass of the object,
v is the velocity of the object.
In this case, the mass of the object is 200 g, which is equivalent to 0.2 kg. So, the equation becomes:
0.02127 J = (1/2)(0.2 kg)v^2
Now, we can solve for the velocity of the object (v):
v^2 = (2 * 0.02127 J) / 0.2 kg
v^2 = 0.2127 J / 0.2 kg
v^2 = 1.0635 m^2/s^2
The square root of the equation gives us:
v ≈ 1.03 m/s
Therefore, the object is moving at approximately 1.03 m/s when it has gained half of the spring's original stored energy.
To find the speed of the object when it has gained half of the spring's original stored energy, we need to use the principle of conservation of mechanical energy. The stored energy in the spring is given by the formula:
E = (1/2)kx^2
Where:
E is the energy stored in the spring
k is the force constant of the spring
x is the displacement of the spring from its equilibrium position
In this case, the spring is compressed 11.0 cm, which is equal to 0.11 meters. The force constant of the spring is given as 35.0 N/cm, which can be converted to 35.0 N/m.
So, the original stored energy in the spring is:
E_initial = (1/2)(35.0 N/m)(0.11 m)^2
Now, we need to find the speed of the object when it has gained half of this energy. We know that the object gains potential energy as it moves towards the equilibrium position, and this potential energy is converted into kinetic energy when the object is released.
The formula for the kinetic energy of an object is given by:
KE = (1/2)mv^2
Where:
KE is the kinetic energy of the object
m is the mass of the object
v is the velocity (speed) of the object
Since the energy gained equals half of the initial stored energy, we can write:
E_gain = (1/2)E_initial
Now we can equate the kinetic energy gained to the energy gain from the spring.
(1/2)mv^2 = (1/2)E_initial
Since the mass of the object is given as 200 g, which is equivalent to 0.2 kg, we can substitute in the values:
(1/2)(0.2 kg)v^2 = (1/2)(35.0 N/m)(0.11 m)^2
Simplifying the equation:
v^2 = (35.0 N/m)(0.11 m)^2
Now we can solve for v by taking the square root of both sides of the equation:
v = √((35.0 N/m)(0.11 m)^2)
Calculating the value:
v ≈ 0.55 m/s
Therefore, the object is moving at approximately 0.55 m/s when it has gained half of the spring's original stored energy.
m = 0.200 kg
k = 3500 N/m
U = (1/2) k x^2 = (1/2)(3500)(0.110)^2
Ke = (1/2) U
(1/2)(.2)v^2 =(1/2)U= (1/4)(3500)(.11)^2