A 750g block is attached to a vertical spring whose stiffness constant is 1N/m . The block is released at the position where the spring is unextended.
a) What is the maximum extension of the spring?
How long does it take the block to reach the lowest point?
I took this course in 1955 and no longer need practice. You are asking basically the same question over and over. I am happy to help but have done enough of them for you.
since it is not extended at the start use y= A cos 2 pi t/T which is A at t = 0 and hits the lowest point, -A at t/T = 1/2
The max extension is 2A down of course, and A=mg/k
To find the maximum extension of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension. The equation for Hooke's Law is:
F = k * x
Where:
F is the force exerted by the spring
k is the stiffness constant of the spring
x is the extension of the spring from its equilibrium position
In this case, the block has a weight (mg) acting downwards, which is balanced by the force exerted by the spring (F). Therefore:
mg = F
Since the weight of the block is given by:
mg = 0.75kg * 9.8m/s^2 = 7.35N
Using Hooke's Law, we can write:
F = k * x
Substituting the values:
7.35N = 1N/m * x
Solving for x, we find:
x = 7.35m
Therefore, the maximum extension of the spring is 7.35m.
To find how long it takes the block to reach the lowest point, we can use the equation of motion for a freely falling object:
s = ut + (1/2)gt^2
Where:
s is the displacement (in this case, the distance the block falls)
u is the initial velocity (which is 0 in this case, as the block is released from rest)
g is the acceleration due to gravity (approximately 9.8m/s^2)
t is the time taken
Since the block is free-falling vertically, the displacement (s) is equal to the maximum extension of the spring (7.35m). Therefore:
7.35 = 0 + (1/2)(9.8)t^2
Simplifying the equation, we find:
(1/2)(9.8)t^2 = 7.35
t^2 = (7.35 * 2) / 9.8
t^2 = 1.5
Taking the square root of both sides, we find:
t = sqrt(1.5)
Therefore, the time it takes for the block to reach the lowest point is approximately 1.22 seconds.
To find the maximum extension of the spring, we can use Hooke's Law, which states that the force applied by a spring is directly proportional to its displacement from its equilibrium position.
The formula for Hooke's Law is: F = -kx
Where:
F = force applied by the spring
k = stiffness constant
x = displacement from the equilibrium position
In this case, the force applied by the spring is equal to the weight of the block, which we can calculate using the formula:
F = m * g
Where:
m = mass of the block
g = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the given values:
m = 750 g = 0.75 kg
g = 9.8 m/s^2
F = 0.75 kg * 9.8 m/s^2
F = 7.35 N
Now, we can rearrange Hooke's Law to solve for x:
x = -F / k
Plugging in the values:
k = 1 N/m
x = -7.35 N / 1 N/m
x = -7.35 m
Since the displacement is negative, the maximum extension of the spring is 7.35 meters.
Next, let's calculate the time taken for the block to reach the lowest point.
To find the time, we can use the following equation of motion:
s = ut + 1/2 * a * t^2
Where:
s = displacement
u = initial velocity
a = acceleration
t = time
In this case, the block is released from an unextended position, so the initial velocity, u, is 0 m/s. The acceleration, a, is equal to the acceleration due to gravity, -g (-9.8 m/s^2).
The displacement is equal to the maximum extension of the spring, which we found to be 7.35 meters.
Plugging in the values:
7.35 = 0 * t + 1/2 * (-9.8) * t^2
Rearranging the equation:
0.5 * (-9.8) * t^2 = 7.35
t^2 = 7.35 / (0.5 * (-9.8))
t^2 = -1.5
Since time cannot be negative, we can ignore the negative value and take the positive square root:
t = √1.5
So, it takes approximately 1.22 seconds for the block to reach the lowest point.