Hooke's law describes a certain light spring of unstretched length 30.0 cm. When one end is attached to the top of a door frame, and a 8.40-kg object is hung from the other end, the length of the spring is 40.00 cm.
(a) Find its spring constant.
____ kN/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 190 N. Find the length of the spring in this situation.
___ m
Hookes law is elongation=kx
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(a) To find the spring constant, we can use Hooke's law equation:
F = k * x
where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring from its original position.
In this case, the spring is stretched from its unstretched length of 30.0 cm to a length of 40.00 cm. The displacement of the spring is:
x = 40.00 cm - 30.0 cm = 10.00 cm = 0.10 m
The force applied to the spring is the weight of the 8.4 kg object hanging from it, given by:
F = m * g
where m is the mass of the object and g is the acceleration due to gravity.
F = 8.4 kg * 9.8 m/s^2
F = 82.32 N
Now we can use Hooke's law to find the spring constant:
82.32 N = k * 0.10 m
Dividing both sides by 0.10 m:
k = 82.32 N / 0.10 m
k = 823.2 N/m
So, the spring constant is 823.2 N/m.
(b) In this situation, two people are pulling in opposite directions on the ends of the spring with a force of 190 N each. The total force applied to the spring is the sum of these two forces:
F = F1 + F2
F = 190 N + 190 N
F = 380 N
We can again use Hooke's law to find the new length of the spring. Rearranging the equation:
F = k * x
we can solve for x:
x = F / k
Substituting the given values:
x = 380 N / 823.2 N/m
x = 0.4618 m
So, the length of the spring in this situation is 0.4618 m.
To solve this problem, we will use Hooke's law, which states that the force exerted by a spring is directly proportional to its deformation (change in length) when subjected to a load.
(a) To find the spring constant (k), we can first calculate the change in length of the spring when the 8.40-kg object is hung from it.
The change in length (Δx) can be calculated using the formula:
Δx = Final length - Initial length
Given:
Initial length (xo) = 30.0 cm = 0.30 m
Final length (xf) = 40.00 cm = 0.40 m
Δx = 0.40 m - 0.30 m
Δx = 0.10 m
Now, we can use Hooke's law formula:
F = k * Δx
where F is the force and Δx is the change in length.
Given:
mass of the object (m) = 8.40 kg
acceleration due to gravity (g) = 9.8 m/s^2 (approximate value)
The force exerted on the spring (F) is equal to the weight of the object, which is given by:
F = m * g
Substituting the values:
F = 8.40 kg * 9.8 m/s^2
F = 82.32 N
Therefore, we can write the equation as:
82.32 N = k * 0.10 m
Now, we can solve for the spring constant (k):
k = 82.32 N / 0.10 m
k = 823.2 N/m
So, the spring constant is 823.2 N/m.
(b) In this situation, two people are pulling on the ends of the spring with forces of 190 N in opposite directions.
Since the forces are equal, the net force on the spring will be zero (190 N - 190 N = 0 N).
According to Hooke's law, when the net force is zero, the spring will be at its equilibrium position, which means the length of the spring will be the same as its unstretched length.
So, the length of the spring in this situation will be 30.0 cm or 0.30 m.