A spring (k = 905 N/m) is hanging from the ceiling of an elevator, and a 4.8-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 = 0.74 m/s2?
To find how much the spring stretches when the elevator is accelerating upward, we need to consider the forces acting on the object attached to the spring.
Let's break down the forces involved:
1. Force due to gravity (Weight): The weight of the object can be calculated using the formula: Weight = mass × acceleration due to gravity. Here, acceleration due to gravity is approximately 9.8 m/s^2. So, Weight = 4.8 kg × 9.8 m/s^2.
2. Force due to the spring (Hooke's Law): According to Hooke's Law, the force exerted by a spring is proportional to the displacement from its equilibrium position. In this case, the displacement of the spring is the amount it stretches or compresses from its original unstrained length. The formula for the force exerted by a spring is: Force = spring constant × displacement. Here, the spring constant (k) is given as 905 N/m.
3. Additional force due to the elevator's acceleration: The acceleration of the elevator (a) causes an additional force on the object, which adds to the force due to gravity. This force can be calculated using Newton's second law: Force = mass × acceleration.
Now, to find the displacement of the spring, we equate the net force on the object to the force exerted by the spring:
Net Force = Force due to gravity + Additional force due to acceleration
mass × acceleration = mass × acceleration due to gravity + spring constant × displacement
Substituting the given values:
4.8 kg × 0.74 m/s^2 = 4.8 kg × 9.8 m/s^2 + 905 N/m × displacement
Now, we can solve for the displacement:
0.74 m/s^2 × 4.8 kg = 9.8 m/s^2 × 4.8 kg + 905 N/m × displacement
3.55 N = 47.04 N + 905 N/m × displacement
3.55 N - 47.04 N = 905 N/m × displacement
-43.49 N = 905 N/m × displacement
displacement = -43.49 N / 905 N/m
displacement ≈ -0.048 m
Hence, the spring stretches by approximately 0.048 meters (or 48 mm) relative to its unstrained length when the elevator is accelerating upward at 0.74 m/s^2. The negative sign indicates that the spring is stretched, opposing the direction of the acceleration.