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?

Fspring = -kx where x is extension up

= -12 x N

m = 0.35 kg

Weight = .35 * 9.8 = 3.45 N

How far will it drop before the force up = the force down? (the equilibrium position)
-k (-x) + m g = 0
12 x = -3.45
x = -.2875
How much potential energy did it lose going from x = 0 to x = -.2875 ?
m g h = 3.45 * .2875 = .992 Joules
that is kinetic energy as it passes the equilibrium point where x = -.2875
(1/2) m v^2 = .992
(1/2) .35 v^2 = .992
v = 2.38 m/s as it passes through equilibrium point
The total energy in the spring increases when it stops at the bottom to start back up> That increase is
.992 J
so the total energy stored in the spring is (1/2) 12 (.2875^2)+ .992
= 1.488 Joules
(1/2)(12) (total extension)^2 = 1.488
total extension = .4980 meters

amplitude of motion = .4980 - .2875 = .2105 meters

b) (1/2) period
where
period = 2 pi sqrt (M/k)
T = 2 pi sqrt(.35/12) = 1.07 seconds
so
0.5 seconds

a) Well, the maximum extension of this spring is like the furthest it is willing to go on a bad hair day - it's all about the stiffness, baby! With a stiffness constant of 12 N/m and a chunky 350 g block, we can use Hooke's law to find the answer. Hooke's law tells us that the force applied by the spring is proportional to its extension. So, we can calculate the force applied on the spring when it's fully extended by using the formula F = kx, where F is the force, k is the stiffness constant, and x is the extension. Since the block is released where the spring is unextended, we know that the force at maximum extension is equal to the weight of the block, which is mg. So, we have mg = kx. Plugging in the values, we get 0.35 kg * 9.8 m/s^2 = 12 N/m * x. Solving for x, we find that the maximum extension of the spring is approximately 2.84 m, which is quite the stretch!

b) Ah, the time it takes for the block to reach the lowest point - the descent that makes roller coasters jealous! To find this, we need to take a small detour into the world of simple harmonic motion. The time period, T, of an object undergoing simple harmonic motion is given by the formula T = 2π√(m/k), where m is the mass of the object and k is the stiffness constant of the spring. Now, we can find the time it takes for the block to reach the lowest point by halving the time period. So, t = T/2 = π√(m/k) = π√(0.35 kg / 12 N/m). Crunching the numbers, we get that the block takes approximately π * 0.228 seconds to reach the lowest point. It's like waiting for the punchline of a joke - it might take a little time, but it's totally worth it!

a) To find the maximum extension of the spring, we need to use the concept of potential energy.

The potential energy of a spring is given by the formula:

Potential energy = (1/2) * k * x^2

where k is the stiffness constant of the spring and x is the extension or compression of the spring from its equilibrium position.

In this case, the block is released from the position where the spring is unextended, which means the extension of the spring is 0 initially.

So, we can write:

Potential energy = (1/2) * k * x^2

At the maximum extension, the potential energy is maximum and the kinetic energy is 0 (as the block momentarily comes to rest before reversing direction).

Hence, the potential energy is equal to the initial gravitational potential energy of the block, which is given by:

Potential energy = m * g * h

where m is the mass of the block, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from which the block is released.

Given that the mass of the block is 350 g (or 0.35 kg) and the stiffness constant of the spring is 12 N/m, we can equate the potential energies to find the maximum extension:

(1/2) * k * x^2 = m * g * h

(1/2) * 12 * x^2 = 0.35 * 9.8 * h

6 * x^2 = 3.43 * h

Since the block is released from the unextended position of the spring, the height of release is zero. Therefore, h = 0.

Hence, the maximum extension of the spring (x) is 0.

b) The time taken for the block to reach the lowest point (when it stops momentarily before reversing its direction) can be determined by considering the periodic motion of the block as it oscillates up and down attached to the spring.

The time period of oscillation (T) can be found using the formula:

T = 2 * π * √(m / k)

where m is the mass of the block and k is the stiffness constant of the spring.

Given that the mass of the block is 350 g (or 0.35 kg) and the stiffness constant of the spring is 12 N/m, we can substitute these values into the formula to find the time period:

T = 2 * π * √(0.35 / 12)

T ≈ 2 * 3.14 * √(0.02917)

T ≈ 2 * 3.14 * 0.1708

T ≈ 1.07 seconds

Therefore, it takes approximately 1.07 seconds for the block to reach the lowest point.

To find the answers to these questions, we can use the concepts of potential energy and simple harmonic motion.

a) To find the maximum extension of the spring, we can use the conservation of energy principle. At the position where the spring is unextended, the block has only gravitational potential energy. At the maximum extension, the block has both gravitational potential energy and elastic potential energy stored in the spring.

The potential energy due to gravity is given by: PE_gravity = m * g * h
Where m is the mass of the block (350 g = 0.35 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height.

To find h, we can use the fact that the block is released from rest at the unextended position, so all the potential energy at that point is converted into the potential energy stored in the spring at the maximum extension.

PE_gravity = PE_spring (at maximum extension)
m * g * h = 0.5 * k * x^2
Where k is the stiffness constant of the spring (12 N/m) and x is the maximum extension we want to find.

Simplifying the equation, we get:
x = sqrt((2 * m * g * h) / k)
x = sqrt((2 * 0.35 kg * 9.8 m/s^2 * h) / 12 N/m)

By substituting values and solving for h, we can find the maximum extension of the spring.

b) To find the time it takes for the block to reach the lowest point, we can use the concept of simple harmonic motion. The block oscillates between two extreme positions while moving up and down.

The period of the motion is given by:
T = 2 * pi * sqrt(m / k)
Where T is the period, m is the mass of the block, and k is the stiffness constant of the spring.

The time it takes for the block to reach the lowest point is half of the period.

t = T / 2

By substituting values and solving for t, we can find the time it takes for the block to reach the lowest point.