Hooke's law describes a certain light spring of unstressed length 36.0 cm. When one end is attached to the top of a door frame and a 6.40 kg object is hung from the other end, the length of the spring is 44.00 cm.
(a) Find its spring constant
I found the correct answer to be k=784 N/m
(b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 170 N. Find the length of the spring in this situation.
I thought the answer was 0.7937m but this is incorrect. I have no other idea on how to solve for part b.
Thank You!
The length of the spring under tension is the amount of stretching PLUS the original unstressed length.
The stretch amount is F/k = 170/784
= 0.217 m = 21.7 cm
Add 36.0 cm to that.
To solve part (b) of the problem, we can use Hooke's law again. Hooke's law states that the force exerted by a spring is proportional to its extension or compression. Mathematically, it is expressed as:
F = -kx
where F is the force applied, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, since two people are pulling in opposite directions, the forces will add up. So the equation becomes:
F_total = -kx
We know that each person is pulling with a force of 170 N, so the total force is:
F_total = 170 N + (-170 N) = 0 N
Plugging this into the equation, we get:
0 = -kx
Since the force is zero, the displacement x must also be zero. This means that the length of the spring will remain unchanged when two people pull on it with equal forces in opposite directions. Therefore, the length of the spring in this situation will still be 36.0 cm.
So the correct answer for part (b) is that the length of the spring remains unchanged at 36.0 cm.
To solve part (a) and find the spring constant, we can use Hooke's law equation:
F = k * Δx
Where:
- F is the force applied to the spring.
- k is the spring constant.
- Δx is the change in length of the spring.
In this case, the spring is initially unstressed and has a length of 36.0 cm (0.36 m). When a 6.40 kg object is hung from it, the length of the spring becomes 44.00 cm (0.44 m).
Since the weight of the object is acting vertically downwards, it creates a force equal to its weight:
F = m * g
Where:
- m is the mass of the object.
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
F = (6.40 kg) * (9.8 m/s^2)
F = 62.72 N
Now we can calculate the change in length of the spring:
Δx = 0.44 m - 0.36 m
Δx = 0.08 m
Plugging the values into Hooke's law equation:
62.72 N = k * 0.08 m
Solving for k:
k = 62.72 N / 0.08 m
k ≈ 784 N/m
So, the spring constant is approximately 784 N/m.
Now, let's move on to part (b).
In part (b), when two people pull in opposite directions on the ends of the spring with forces of 170 N each, the net force on the spring is the sum of the two forces:
F_net = F1 + F2
Where:
- F_net is the net force on the spring.
- F1 and F2 are the forces applied by each person.
Since the forces are pulling in opposite directions, the net force will be the difference between the two forces:
F_net = F1 - F2
In this case, both forces are 170 N:
F_net = 170 N - 170 N
F_net = 0 N
Since the net force is zero, the spring will be in its unstressed state, and its length will be equal to its unstressed length of 36.0 cm (0.36 m).
Therefore, the length of the spring in this situation is 0.36 m.