When a 3.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.07 cm.
(a) What is the force constant of the spring?
N/m
(b) If the 3.00-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?
cm
(c) How much work must an external agent do to stretch the same spring 9.10 cm from its unstretched position?
J
k=force/stretch= 3g/.0307 in N/m
b. x=force/k=1.5/k I bet it is half of 3.07
c. work= 1/2 k x^2 watch units, x has to be in m.
To answer the given questions, we need to apply Hooke's law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position.
(a) The force constant of the spring, also known as the spring constant (k), can be determined using Hooke's law formula:
F = -kx
Where:
F is the force applied on the spring (in N)
k is the spring constant (in N/m)
x is the displacement or stretch of the spring (in m)
In this case, the given displacement is 3.07 cm, which is equivalent to 0.0307 m. The force (F) can be calculated using the weight equation:
F = mg
Where:
m is the mass of the object (in kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
For the given 3.00 kg object, the force can be calculated as:
F = (3.00 kg) * (9.8 m/s^2) = 29.4 N
Now, substituting the values into the Hooke's law formula:
29.4 N = -k * 0.0307 m
Rearranging the equation to solve for k:
k = -29.4 N / 0.0307 m
Calculating k:
k ≈ -957 N/m (Note: The negative sign indicates that the spring is being stretched.)
Therefore, the force constant of the spring is approximately 957 N/m.
(b) To determine how far the spring will stretch when a 1.50 kg block is hung on it, we can again use Hooke's law:
F = -kx
We know the spring constant (k) from the previous calculation, which is approximately 957 N/m. The force (F) can be calculated using the weight equation:
F = mg
For the 1.50 kg block:
F = (1.50 kg) * (9.8 m/s^2) = 14.7 N
Substituting the values into the Hooke's law formula:
14.7 N = -957 N/m * x
Solving for x:
x = 14.7 N / -957 N/m
Calculating x:
x ≈ -0.0153 m (Note: Again, the negative sign indicates that the spring is being stretched.)
To convert the displacement to centimeters, multiply by 100:
x ≈ -0.0153 m * 100 = -1.53 cm
Therefore, when a 1.50 kg block is hung on the spring, it will stretch by approximately 1.53 cm.
(c) To calculate the work required to stretch the spring 9.10 cm from its unstretched position, we can use the formula for work:
Work (W) = (1/2) * k * x^2
Where:
W is the work done on the spring (in J)
k is the spring constant (in N/m)
x is the displacement or stretch of the spring (in m)
Given that the displacement (x) is 9.10 cm, which is equivalent to 0.091 m, and the spring constant (k) is 957 N/m:
W = (1/2) * 957 N/m * (0.091 m)^2
Calculating W:
W ≈ (1/2) * 957 N/m * (0.008281 m^2)
W ≈ 3.98 J (rounded to two decimal places)
Therefore, the external agent must do approximately 3.98 J of work to stretch the same spring 9.10 cm from its unstretched position.
To solve these physics problems, we'll use Hooke's law: F = kx, where F is the force, k is the force constant of the spring, and x is the displacement from the equilibrium position.
(a) To find the force constant of the spring, we need to calculate it using the given information. The force exerted by the spring is equal to the weight of the object, so we have:
F = mg, where m = 3.00 kg and g = 9.8 m/s^2 (acceleration due to gravity).
We also know that the displacement is x = 3.07 cm = 0.0307 m.
Substituting these values into Hooke's law, we have:
mg = kx
k = (mg) / x
k = (3.00 kg)(9.8 m/s^2) / 0.0307 m
k ≈ 973 N/m
Therefore, the force constant of the spring is approximately 973 N/m.
(b) Now, let's determine how far the spring will stretch when a 1.50-kg block is hung on it. Since we know the force constant of the spring (k) and the mass of the object (m), we can calculate the displacement as follows:
F = mg = (1.50 kg)(9.8 m/s^2) = 14.7 N
Substituting these values into Hooke's law, we have:
F = kx
x = F / k
x = 14.7 N / 973 N/m
x ≈ 0.0151 m = 1.51 cm
Therefore, when a 1.50-kg block is hung on the spring, it will stretch approximately 1.51 cm.
(c) Finally, we need to calculate the work done to stretch the same spring 9.10 cm from its unstretched position. The work done can be determined using the formula:
W = (1/2) kx^2
where W is the work done, k is the force constant of the spring, and x is the displacement.
Given that the displacement x = 9.10 cm = 0.091 m, and the force constant k ≈ 973 N/m (calculated earlier), we can calculate the work done:
W = (1/2) (973 N/m) (0.091 m)^2
W ≈ 0.394 J
Therefore, the external agent must do approximately 0.394 Joules of work to stretch the spring 9.10 cm from its unstretched position.