A 0.56-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.13 to 0.26 m (relative to its unstrained length), the speed of the sphere decreases from 5.45 to 4.90 m/s. What is the spring constant of the spring? (The sphere hangs from the bottom of the spring.)
Answer in N/m
To solve this problem, we can use the formula for the potential energy of a spring:
Potential energy (PE) = 0.5 * k * x^2
where k is the spring constant and x is the displacement from the equilibrium position.
First, we need to calculate the initial and final potential energies of the spring. Since the sphere is at rest at the extended position, the initial potential energy (PEi) is zero. The final potential energy (PEf) can be calculated using the mass and speed of the sphere at that position.
PEf = 0.5 * m * v^2
Given:
Mass of the sphere (m) = 0.56 kg
Initial speed (v_i) = 5.45 m/s
Final speed (v_f) = 4.90 m/s
Displacement (x) = 0.26 m - 0.13 m = 0.13 m
Using the potential energy formula and substituting the values, we have:
PEf = 0.5 * 0.56 kg * (4.90 m/s)^2
Next, we need to calculate the difference in potential energy (ΔPE) between the initial and final positions:
ΔPE = PEf - PEi
Since PEi is zero, ΔPE = PEf.
Now, let's substitute the values and calculate ΔPE:
ΔPE = 0.5 * 0.56 kg * (4.90 m/s)^2
Now, we have the change in potential energy. We can relate ΔPE to the spring constant (k) using Hooke's Law:
ΔPE = 0.5 * k * x^2
Substituting the known values, we can solve for k:
0.5 * k * (0.13 m)^2 = 0.5 * 0.56 kg * (4.90 m/s)^2
Let's calculate it:
k * (0.13 m)^2 = (0.56 kg * (4.90 m/s)^2)
Simplifying the equation:
k * 0.0169 m^2 = 11.0464 N
Finally, divide both sides of the equation by 0.0169 m^2 to solve for k:
k = 11.0464 N / 0.0169 m^2
After performing the division, we find:
k ≈ 652.54 N/m
Therefore, the spring constant of the spring is approximately 652.54 N/m.
To find the spring constant, we can use the equation for the potential energy of a spring:
Potential Energy (PE) = (1/2) * k * x^2
Where:
PE is the potential energy of the spring
k is the spring constant
x is the displacement of the spring from its equilibrium position
In this case, we have two points with different speeds and displacements. Let's use the two points to find the spring constant.
At the first point:
Displacement (x1) = 0.26 m
Speed (v1) = 5.45 m/s
At the second point:
Displacement (x2) = 0.13 m
Speed (v2) = 4.90 m/s
First, let's find the change in potential energy (ΔPE):
ΔPE = PE2 - PE1
ΔPE = (1/2) * k * x2^2 - (1/2) * k * x1^2
ΔPE = (1/2) * k * (x2^2 - x1^2)
ΔPE = (1/2) * k * (0.13^2 - 0.26^2)
ΔPE = (1/2) * k * (0.0169 - 0.0676)
ΔPE = (1/2) * k * (-0.0507)
Now, let's find the change in kinetic energy (ΔKE):
ΔKE = KE2 - KE1
ΔKE = (1/2) * m * v2^2 - (1/2) * m * v1^2
ΔKE = (1/2) * 0.56 kg * (4.90^2 - 5.45^2)
ΔKE = (1/2) * 0.56 kg * (24.01 - 29.70)
ΔKE = (1/2) * 0.56 kg * (-5.69)
ΔKE = -1.98 J
Since kinetic energy decreases, ΔKE is negative.
According to the conservation of energy, ΔPE + ΔKE = 0
-1.98 J + (1/2) * k * (-0.0507) J = 0
Simplifying the equation:
k * (-0.0507) = 1.98
k = 1.98 / (-0.0507)
k ≈ -39.00 N/m
However, since the spring constant should have a positive value, we take the magnitude of the result:
k ≈ | -39.00 | ≈ 39.00 N/m
Therefore, the spring constant of the spring is approximately 39.00 N/m.