A 63 kg student is standing atop a spring in an elevator that is accelerating upward at 2.5 m/s^2 . The spring constant is 2700 N/m.
By how much is the spring compressed?
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Well, you could say the student is really springing into action in this elevator! Now, to determine how much the spring is compressed, we can use Hooke's Law. According to Hooke's Law, the force exerted by the spring (F) is equal to the spring constant (k) multiplied by the displacement (x).
In this case, the force exerted by the spring is equal to the weight of the student, which is given by the equation F = mg, where m is the mass of the student and g is the acceleration due to gravity.
So, let's calculate the force exerted by the spring:
F = mg
F = (63 kg)(9.8 m/s^2)
F ≈ 617.4 N
Now, we can use Hooke's Law:
F = kx
Rearranging the equation, we get:
x = F / k
x = 617.4 N / 2700 N/m
x ≈ 0.229 m
So, the spring is compressed by approximately 0.229 meters. Don't worry, the student will bounce back up!
To determine the amount of compression in the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement or compression of the spring.
First, let's find the net force acting on the student. We can calculate this by using Newton's second law of motion, which states that the net force is equal to the mass multiplied by the acceleration.
Net force = mass × acceleration
The mass of the student is given as 63 kg, and the acceleration of the elevator is 2.5 m/s^2. Plugging these values into the equation:
Net force = 63 kg × 2.5 m/s^2
Net force = 157.5 N
Since the force exerted by the spring is equal to the net force, we can write:
Force = spring constant × compression
The spring constant is given as 2700 N/m. Rearranging the equation to solve for compression:
Compression = Force / spring constant
Substituting the values:
Compression = 157.5 N / 2700 N/m
Compression ≈ 0.0583 m
Therefore, the spring is compressed by approximately 0.0583 meters.
m(g+a)=kx
x=m(g+a)/k