when a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm. A) what is the force constant of the spring. B) if the 2.50 kg object is removed how far will the spring strecth if a 1.25 kg block is hung on it. C) how much work must an external agent do to strecth the same spring 8.00cm from its unstreched position.

In your "C" answer, second line, you should have used 0.08m, not 0.0888m (that was the K constant).

Therfore, the result is W=Fd= 71*0.08= 5.68 Joules

2.84J

A) Well, to find the force constant of the spring, you need to know Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement. In equation form, F = kx, where F is the force applied, k is the force constant, and x is the displacement. In this case, you have the mass of the object (2.50 kg) and the displacement (2.76 cm). However, we need to convert the displacement into meters for the equation to work. So, 2.76 cm is equivalent to 0.0276 m. Now, we can plug in the values: F = kx. Rearranging the equation to solve for k, we get k = F/x. Since the weight of the object is equal to the force applied, F = mg, where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, k = (2.50 kg * 9.8 m/s^2) / 0.0276 m. Crunching the numbers, the force constant of the spring is about 891.30 N/m.

B) Now, if we remove the 2.50 kg object and hang a 1.25 kg block on the spring, we can find the new displacement. We know the force constant of the spring (891.30 N/m). Using Hooke's Law again, F = kx, we can now solve for x, but first we need to find the force applied. The force applied can be calculated as the weight of the block, which is F = mg = (1.25 kg * 9.8 m/s^2). Plugging in the values, k = F/x. Rearranging the equation to solve for x, we get x = F/k. Crunching the numbers again, the new displacement of the spring with the 1.25 kg block is approximately 0.0699 m.

C) Lastly, to find the work done to stretch the spring 8.00 cm from its unstretched position, we'll use the formula for work: W = (1/2)kx^2. We know the force constant of the spring (891.30 N/m) and the new displacement (0.08 m, since we need to convert 8.00 cm into meters). Plugging these values into the work formula, we find W = (1/2)(891.30 N/m)(0.08 m)^2. Crunching the numbers one more time, the work required to stretch the spring 8.00 cm is approximately 2.27 J.

Well, that was quite a journey of spring-related math! I hope my humor made it a bit more bearable.

A) To find the force constant of the spring, you can use Hooke's Law equation: F = kx, where F is the force applied, k is the force constant, and x is the displacement of the spring.

From the given information, the mass of the object is 2.50 kg and the spring stretches 2.76 cm (or 0.0276 m). We need to convert the mass to force using the equation F = mg, where g is the acceleration due to gravity (approximately 9.8 m/s²).

So, F = (2.50 kg) * (9.8 m/s²) = 24.5 N (Newtons)

Now, applying Hooke's Law:

F = kx
24.5 N = k * 0.0276 m

Solving for k, we get:

k = 24.5 N / 0.0276 m ≈ 887.7 N/m

Therefore, the force constant of the spring is approximately 887.7 N/m.

B) To find how far the spring will stretch when a 1.25 kg block is hung on it, we can use the same equation F = kx and rearrange it to solve for x:

x = F / k

We have already calculated the force constant, which is 887.7 N/m. Now we need to find the force when a 1.25 kg block is hung on the spring:

F = (1.25 kg) * (9.8 m/s²) = 12.25 N

Plugging the values into the equation:

x = 12.25 N / 887.7 N/m ≈ 0.0138 m

Therefore, the spring will stretch approximately 0.0138 m (or 1.38 cm) when the 1.25 kg block is hung on it.

C) To calculate the work done to stretch the spring by 8.00 cm, we can use the formula for work:

Work = 0.5 * k * (x² - xi²)

In this formula, k is the force constant of the spring, x is the final displacement, and xi is the initial displacement (0 in this case since the spring is initially unstretched).

Given information: x = 8.00 cm = 0.08 m, k = 887.7 N/m

Plugging the values into the equation:

Work = 0.5 * 887.7 N/m * (0.08² - 0²)
= 0.5 * 887.7 N/m * (0.0064 m²)
≈ 2.84 J (Joules)

Therefore, an external agent must do approximately 2.84 Joules of work to stretch the spring 8.00 cm from its unstretched position.

Wo = mg = 2.50kg * 9.8N/kg = 24.5N. =

Weight of objedt.

A. Fk = Wo / d = 24.5 / 2.76 = 8.88N/cm

B. d=(1.25kg/2.50kg) * 2.76cm = 1.38cm

C. F = 8.88N/cm * 8cm = 71N.
W = Fd = 71 * 0.0888m = 6.31 Joules.