1. A 5 kg weight is hung from a vertical spring. The spring stretches by 5 cm. How much mass should be hung from the spring so it will stretch by 10 cm from its original length?

A. 15 kg
B. 10 kg
C. 20 kg
D. 25 kg

2. What is the spring constant of a spring that is composed by 0.250 cm when a force of 35 N is applied?
A. 140 N/cm
B. 100 N/cm
C. 200 N/cm
D. 150 N/cm

3. Spring A has a spring constant of 5 N/m and spring B has a force constant of 9 N/m. A weight of 2 kg is applied to spring A and a weight of 10 kg is applied to spring B. Which is true for the final displacements of springs A and B?
A. Spring B has a larger displacement than spring A.
B. Spring A has a larger displacement than spring B.
C. Spring A and spring B have the same displacement.

Help!!!

1. C - 12.8

2. B - 32 N/m
3. A - 31 cm
4. A - Spring B has a larger displacement than spring A.
5. C - 0.456 N
6. B - 10 kg
7. B - 82.4 cm
8. C - 25 N/m
9. A - 25 N
10. B - 86 m/s^2

1. To determine how much mass should be hung from the spring to stretch it by 10 cm, we can use Hooke's Law which states that the force exerted by a spring is proportional to the displacement it undergoes. The formula for Hooke's Law is F = kx, where F is the force, k is the spring constant, and x is the displacement.

First, let's find the spring constant (k) using the given information. Since the spring stretches by 5 cm when a 5 kg weight is hung from it, we can write the equation F = kx as 5 kg * 9.8 m/s^2 = k * 0.05 m. Solving for k, we get k = (5 kg * 9.8 m/s^2) / 0.05 m = 980 N/m.

Now, we can use this spring constant to find the mass (m) needed to stretch the spring by 10 cm. Again, using F = kx, we substitute the given displacement of 0.10 m and solve for mass: F = (980 N/m)(0.10 m) = m * 9.8 m/s^2. Solving for m, we get m = (980 N/m * 0.10 m) / 9.8 m/s^2 = 10 kg.

Therefore, the correct answer is B. 10 kg.

2. To find the spring constant of a spring, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is proportional to the displacement it undergoes. The formula for Hooke's Law is F = kx, where F is the force, k is the spring constant, and x is the displacement.

In this case, the displacement is given as 0.250 cm (which is equivalent to 0.0025 m), and the force applied is 35 N. Using the formula F = kx, we can solve for the spring constant: k = F / x = 35 N / 0.0025 m = 14,000 N/m.

Therefore, the correct answer is A. 140 N/cm (since 1 m = 100 cm, the conversion can be done by multiplying with 100).

3. To compare the final displacements of springs A and B, we can use Hooke's Law which states that the force exerted by a spring is proportional to the displacement it undergoes. The formula for Hooke's Law is F = kx, where F is the force, k is the spring constant, and x is the displacement.

For spring A, which has a spring constant of 5 N/m, and a weight of 2 kg applied to it, we can calculate the displacement using F = kx, where F = 2 kg * 9.8 m/s^2 and k = 5 N/m. Plugging in the values, we get 2 kg * 9.8 m/s^2 = 5 N/m * x_A. Solving for x_A, we find x_A = (2 kg * 9.8 m/s^2) / (5 N/m) = 3.92 m.

For spring B, which has a spring constant of 9 N/m, and a weight of 10 kg applied to it, we can use the same formula to calculate the displacement. Plugging in the values, we get 10 kg * 9.8 m/s^2 = 9 N/m * x_B. Solving for x_B, we find x_B = (10 kg * 9.8 m/s^2) / (9 N/m) = 10.88 m.

Therefore, the correct answer is A. Spring B has a larger displacement than spring A.

1. To determine the mass that should be hung from the spring to stretch it by 10 cm, you can use the formula for the differential equation of a spring:

F = k * x

Where F is the force exerted on the spring, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, a 5 kg weight stretches the spring by 5 cm, so the force exerted on the spring can be calculated as:

F1 = m1 * g = 5 kg * 9.8 m/s^2 = 49 N

Given that the displacement for this case is 5 cm, we can now calculate the spring constant k for the current setup:

F1 = k * x1
49 N = k * 0.05 m

Solving for k, we find:
k = 49 N / 0.05 m = 980 N/m

Now, we want to find the mass that should be hung from the spring to stretch it by 10 cm. Let's call this mass m2.

Using the formula again, we can set up the equation:

F2 = m2 * g = k * x2

We know that F2 is equal to the force exerted by the 5 kg weight, which is 49 N. We also know that x2 is 0.1 m. Plugging in these values, we can solve for m2:

49 N = m2 * 9.8 m/s^2 = 980 N/m * 0.1 m
49 N = 98 N * m2
m2 = 49 N / 98 N = 0.5 kg

Therefore, the mass that should be hung from the spring to stretch it by 10 cm is 0.5 kg.

The correct answer is not provided in the given options.

1. Mass = 10cm/5cm * 5kg =

2. k = 35/0.25cm =

3. M*g = 2 * 9.8 = 19.6 N. applied.
Spring A: Disp. = 19.6N * 1m/5N = 3.92 m.

M*g = 10 * 9.8 = 98 N. applied.
Spring B: Disp. = 98N * 1m/9N = 10.9 m.