A 350 g block is attached to a vertical spring whose stiffness constant is 12 N/m . The block is released at the position where the spring is unextended.
a) What is the maximum extension of the spring?
b)How long does it take the block to reach the lowest point?
To find the maximum extension of the spring (a), we can use the concept of potential energy and Hooke's Law.
a) The potential energy stored in a spring is given by the formula:
Potential Energy = (1/2) * k * x^2
Where:
k = stiffness constant (12 N/m)
x = extension of the spring
Since the block is released from the unextended position, the potential energy is converted solely into the potential energy of the spring. At the maximum extension of the spring, the potential energy of the block is zero. Therefore:
Potential Energy = (1/2) * k * x^2 = 0
Solving this equation, we find that x = 0.
Hence, the maximum extension of the spring is 0.
b) To find how long it takes the block to reach the lowest point, we need to calculate the time taken for one complete oscillation. The time period of an oscillation for a mass-spring system is given by the formula:
Time period (T) = 2π * √(m/k)
Where:
m = mass of the block (350 g = 0.35 kg)
k = stiffness constant (12 N/m)
Plugging in the values, we get:
T = 2π * √(0.35/12)
Calculating this equation, we find the time period (T) for one complete oscillation.
Since the time taken to reach the lowest point is half the time period, we can divide the time period by 2 to get the answer.
Therefore, the time taken for the block to reach the lowest point is T/2.
To find the maximum extension of the spring, we need to consider the potential energy of the block-spring system.
a) The potential energy stored in a spring can be calculated using the formula:
Potential energy = (1/2) k x^2
Where k is the stiffness constant of the spring (k = 12 N/m) and x is the extension of the spring.
At the maximum extension, the potential energy is equal to the gravitational potential energy of the block:
Potential energy = mgh
Where m is the mass of the block (m = 350 g = 0.35 kg), g is the acceleration due to gravity (g = 9.8 m/s^2), and h is the maximum extension of the spring.
Setting the two equations equal to each other, we have:
(1/2) k x^2 = mgh
Plugging in the given values, we can solve for x:
(1/2) * 12 * x^2 = 0.35 * 9.8 * h
6x^2 = 3.43h
x^2 = 0.572h
To find the maximum extension, we need to find the value of x when h is maximum. Since h is maximum when x is maximum, we can rearrange the equation:
x^2 = 0.572h
x = sqrt(0.572h)
We know that when the spring is unextended, x = 0, so we can solve for h:
0 = sqrt(0.572h)
0.572h = 0
Therefore, there is no maximum extension of the spring.
b) To find the time it takes for the block to reach the lowest point, we can use the equation for the oscillation period:
T = 2π sqrt(m/k)
Where m is the mass of the block (m = 350 g = 0.35 kg) and k is the stiffness constant of the spring (k = 12 N/m).
Plugging in the given values, we can calculate T:
T = 2π sqrt(0.35 / 12)
T ≈ 2π sqrt(0.029167)
T ≈ 2π * 0.170924
T ≈ 1.0738 s
Therefore, it takes approximately 1.0738 seconds for the block to reach the lowest point.