A spring (k = 857 N/m) is hanging from the ceiling of an elevator, and a 5.9 kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at a = 0.81 m/s2?
force= kx
mg+ma=kx
solve for a.
Use kx=mg+ma
857x = (5.9*9.8) + (5.9*.81)
thus, x= .073
Why did the spring become an elevator repairman? Because it wanted to have a "spring" in its step!
Now, let's solve this problem.
We know that the force due to gravity (mg) and the force due to the acceleration of the elevator (ma) must be balanced by the force exerted by the spring (kx).
So, we can write the equation as follows:
mg + ma = kx
Plugging in the given values:
(5.9 kg)(9.8 m/s^2) + (5.9 kg)(0.81 m/s^2) = (857 N/m) x
Now, let's solve for x, the stretch of the spring relative to its unstrained length:
(5.9 kg)(9.8 m/s^2 + 0.81 m/s^2) = (857 N/m) x
57.82 N = 857 N/m x
Dividing both sides by 857 N/m:
x = 0.06751 m
So, the spring stretches by approximately 0.06751 meters relative to its unstrained length when the elevator is accelerating upward at 0.81 m/s^2.
To find the stretch of the spring, we need to find the net force acting on the object and divide it by the spring constant (k).
The equation for the net force is:
Net Force = ma + mg
where:
m = mass of the object = 5.9 kg
a = acceleration of the elevator = 0.81 m/s^2
g = acceleration due to gravity = 9.8 m/s^2
Substituting the given values into the equation:
Net Force = (5.9 kg)(0.81 m/s^2) + (5.9 kg)(9.8 m/s^2)
Net Force = 4.779 kg·m/s^2 + 57.62 kg·m/s^2
Net Force = 62.399 kg·m/s^2
Next, we can find the stretch of the spring (x) by dividing the net force by the spring constant (k):
x = Net Force / k
x = 62.399 kg·m/s^2 / 857 N/m
x ≈ 0.0727 m
Therefore, the spring stretches by approximately 0.0727 meters relative to its unstrained length when the elevator is accelerating upward at 0.81 m/s^2.
To find the stretch of the spring, we need to use Newton's second law and the formula for the force exerted by a spring.
1. Start by calculating the weight force acting on the object:
F_g = m * g
where m is the mass of the object (5.9 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_g = 5.9 kg * 9.8 m/s^2 = 57.82 N
2. Next, calculate the net force acting on the object:
F_net = m * a
where a is the acceleration of the elevator (0.81 m/s^2).
F_net = 5.9 kg * 0.81 m/s^2 = 4.779 N
3. The force exerted by the spring is given by Hooke's Law:
F_spring = k * x
where k is the spring constant (857 N/m) and x is the amount the spring stretches or compresses.
4. Since the object is connected to the spring, the net force acting on the object is the sum of the gravitational force and the force exerted by the spring:
F_net = F_g + F_spring
5. Set up the equation by substituting the values we calculated:
4.779 N = 57.82 N + 857 N/m * x
6. Solve for x:
4.779 N = 57.82 N + 857 N/m * x
4.779 N - 57.82 N = 857 N/m * x
-53.041 N = 857 N/m * x
7. Divide both sides of the equation by 857 N/m to isolate x:
x = (-53.041 N) / (857 N/m)
x ≈ -0.062 m
Therefore, the spring stretches by approximately 0.062 meters (or 6.2 cm) relative to its unstrained length when the elevator is accelerating upward at 0.81 m/s^2. Note that the negative sign indicates that the spring is stretched upwards.