When a 2.60-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.30 cm.

(a) What is the force constant of the spring?
(b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it?
(c) How much work must an external agent do to stretch the same spring 8.40 cm from its unstretched position?

2.30 cm = .0230 meters

2.60 kg * 9.8 m/s^2 = 25.5 Newtons
so
k = 25.5/.0230 = 1108 N/m

1.30*9.8 = 1108 x
x = .0115 m (half as far of course)

Energy stored = work done = (1/2) k x^2
= .5*1108*(.0840)^2

(a) Well, well, well, looks like we have a spring determining its own fate! To find the force constant, we can use Hooke's law, which states that the force constant is equal to the force divided by the displacement.

So, we have an object with a mass of 2.60 kg hanging on the spring and stretching it by 2.30 cm. First, we need to convert the displacement to meters: 2.30 cm = 0.023 m. Now, we know that the force acting on the spring is equal to the weight of the object, which is m*g, where m is the mass of the object and g is the acceleration due to gravity.

The force constant, k, can be calculated by dividing the force (m*g) by the displacement (0.023 m). So, we have k = (2.60 kg * 9.8 m/s^2) / 0.023 m.

Now let's calculate that and make the force constant happy!

(b) Oh, the poor spring! Once the 2.60-kg object is removed, another 1.30-kg block comes along to take its place. How far will the spring stretch? Let's find out!

Since the spring is light enough to not affect the current object's mass, we can simply consider the new mass and calculate the displacement using the same force constant, k.

We can use Hooke's law once again and rearrange the formula to solve for the displacement:

displacement = (force / k)

The force acting on the spring is now the weight of the 1.30-kg block, which is m*g. So, we have displacement = (m*g / k).

Plug in the values and let me do the math for you!

(c) Stretching the spring even further, huh? Looks like someone likes a challenge! To calculate the work done to stretch the same spring by 8.40 cm, we can use the formula:

work = (1/2) * k * (displacement^2)

But hold on! We can't just go ahead and plug in the displacement as 8.40 cm because that little rascal needs to be in meters. So, let's convert the displacement to meters: 8.40 cm = 0.084 m.

Now we're ready to calculate the work done! Plug in the values into the formula, and voila!

(a) To find the force constant of the spring, we can use Hooke's law equation:

F = -kx

where F is the force applied on the spring, k is the force constant, and x is the displacement of the spring.

Given:
Mass of the object, m = 2.60 kg
Displacement of the spring, x = 2.30 cm = 0.0230 m

At equilibrium, the weight of the object is balanced by the force exerted by the spring:

mg = kx

Substituting the given values:

k = (m * g) / x

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging in the values:

k = (2.60 kg * 9.8 m/s^2) / 0.0230 m
k ≈ 1116.52 N/m

Therefore, the force constant of the spring is approximately 1116.52 N/m.

(b) To find how far the spring will stretch when a 1.30-kg block is hung on it, we can use the force constant we calculated in part (a) and Hooke's law:

F = -kx

Solving for x:

x = -F / k

The force applied on the spring when the 1.30-kg block is hung on it can be calculated using the equation:

F = mg

Given:
Mass of the block, m = 1.30 kg

F = (1.30 kg * 9.8 m/s^2) = 12.74 N

Plugging the values into the equation:

x = -(12.74 N) / 1116.52 N/m
x ≈ -0.0114 m

Note that the negative sign indicates the direction of the displacement, i.e., the spring compresses by 0.0114 m.

Therefore, the spring will stretch approximately 0.0114 m (or 1.14 cm) when the 1.30-kg block is hung on it.

(c) To calculate the work done to stretch the spring by 8.40 cm, we can use the formula:

Work = (1/2) * k * (x^2)

Given:
Displacement, x = 8.40 cm = 0.0840 m

Plugging in the values:

Work = (1/2) * (1116.52 N/m) * (0.0840 m)^2
Work ≈ 3.37 J

Therefore, an external agent must do approximately 3.37 J of work to stretch the same spring from its unstretched position by 8.40 cm.

To find the answers to these questions, we can use Hooke's law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position.

(a) To find the force constant of the spring (k), we can use the formula:

k = F / x

where F is the force applied to the spring and x is the displacement. In this case, the force applied is the weight of the object, which is equal to its mass multiplied by the acceleration due to gravity (g = 9.8 m/s^2). The displacement x is given as 2.30 cm (or 0.0230 m). So we have:

F = m * g
F = 2.60 kg * 9.8 m/s^2
F = 25.48 N

Now we can calculate the force constant:

k = F / x
k = 25.48 N / 0.0230 m
k ≈ 1109.57 N/m

(b) To find the amount the spring will stretch when a 1.30-kg block is hung on it, we can use the same formula:

F = k * x

where F is the force applied, k is the force constant of the spring, and x is the displacement. However, in this case, we need to solve for x. We know that the force applied is the weight of the block, which is equal to its mass multiplied by the acceleration due to gravity. Let's denote the new displacement as x₂ and the new mass as m₂. We can set up the following equation:

m₂ * g = k * x₂

Solving for x₂, we get:

x₂ = (m₂ * g) / k

Plugging in the given values, we have:

x₂ = (1.30 kg * 9.8 m/s^2) / 1109.57 N/m
x₂ ≈ 0.0114 m or 1.14 cm

Therefore, the spring will stretch approximately 1.14 cm when the 1.30-kg block is hung on it.

(c) To find the work required to stretch the spring 8.40 cm from its unstretched position, we need to calculate the potential energy stored in the spring. The potential energy stored in a spring is given by the equation:

U = (1/2) * k * x^2

where U is the potential energy, k is the force constant of the spring, and x is the displacement. To find the work done, which is the same as the potential energy stored, we can use this equation.

We are given the displacement x = 8.40 cm (or 0.0840 m) and the force constant k = 1109.57 N/m. Plugging in these values, we have:

U = (1/2) * 1109.57 N/m * (0.0840 m)^2
U ≈ 4.905 J

Therefore, an external agent must do approximately 4.905 Joules of work to stretch the spring 8.40 cm from its unstretched position.