By how much will a 1.52 kg object stretch a spring that is suspended vertically if the spring constant is 135 N/m?


The object is pulled straight down by an additional distance of 25.4 cm and released from rest. Find the speed with which the object passes through its original motionless position from the previous question on the way up?

To find the amount the spring stretches when a 1.52 kg object is suspended from it, we can use Hooke's Law:

F = k * x

Where:
F = force applied to the spring
k = spring constant
x = displacement of the spring

First, let's find the force applied to the spring:

F = m * g

Where:
m = mass of the object
g = acceleration due to gravity

m = 1.52 kg
g = 9.8 m/s^2

F = 1.52 kg * 9.8 m/s^2
F = 14.896 N

Now, let's find the displacement of the spring:

F = k * x

x = F / k

k = 135 N/m

x = 14.896 N / 135 N/m
x = 0.11 m

Therefore, the spring will stretch by 0.11 meters.

To find the speed with which the object passes through its original motionless position on the way up, we can use conservation of energy.

The potential energy at the bottom when the object is released is given by:

PE_initial = m * g * h_initial

Where:
m = mass of the object
g = acceleration due to gravity
h_initial = initial displacement of the object

m = 1.52 kg
g = 9.8 m/s^2
h_initial = 25.4 cm = 0.254 m

PE_initial = 1.52 kg * 9.8 m/s^2 * 0.254 m
PE_initial = 3.7636 J

At the highest position, when the object momentarily comes to rest, all the potential energy is converted to kinetic energy.

KE_final = 1/2 * m * v^2

Where:
m = mass of the object
v = velocity of the object

KE_final = 3.7636 J

Rearranging the equation, we get:

v = sqrt(2 * KE_final / m)

v = sqrt(2 * 3.7636 J / 1.52 kg)
v ≈ 1.97 m/s

Therefore, the velocity with which the object passes through its original motionless position on the way up is approximately 1.97 m/s.

To find the stretch of the spring caused by the 1.52 kg object, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement or stretch of the spring.

The formula for Hooke's Law is:

F = k * x,

where F represents the force exerted by the spring, k is the spring constant (135 N/m in this case), and x is the displacement or stretch of the spring.

First, let's find the force exerted by the spring when the 1.52 kg object is suspended from it:

F = k * x,
F = 135 N/m * x,

The object is initially at rest, so the force exerted by the spring must balance the weight of the object. The weight of the object is given by:

weight = mass * acceleration due to gravity,
weight = 1.52 kg * 9.8 m/s^2.

Since the object is at rest, the force exerted by the spring must be equal to the weight of the object:

135 N/m * x = 1.52 kg * 9.8 m/s^2.

Now we can solve for x, the stretch of the spring:

x = (1.52 kg * 9.8 m/s^2) / 135 N/m.

This will give us the stretch of the spring caused by the object. Now, let's move on to the second part of the question.

The object is pulled an additional distance of 25.4 cm and released from rest. We want to find the speed of the object as it passes through its original motionless position on its way up.

Since the object is released from rest, we can use the principle of conservation of energy to solve this problem.

The initial potential energy of the object is given by:

PE_initial = (1/2) * k * x^2,

where k is the spring constant and x is the stretch of the spring.

The final kinetic energy of the object is given by:

KE_final = (1/2) * m * v^2,

where m is the mass of the object and v is the speed of the object.

According to the principle of conservation of energy, the initial potential energy must be equal to the final kinetic energy:

PE_initial = KE_final.

Substituting the formulas for potential and kinetic energy:

(1/2) * k * x^2 = (1/2) * m * v^2.

Now we can solve for v, the speed of the object:

v = sqrt((k * x^2) / m).

Substituting the values given in the problem:

v = sqrt((135 N/m * (0.254 m)^2) / 1.52 kg).

This will give us the speed with which the object passes through its original motionless position on its way up.